STRUCTURE OF ATOM This Chapter “Structure of Atom” is taken from our:
ISBN : 9789386146533
1.
Which is not correct amongst the following statements about cathode rays :
6.
The angular momentum of the electron in first excited energy state of hydrogen atom is
(a) They are deflected by electric and magnetic fields (b) e/m ratio of the particles forming them depends upon the gas filled in the discharge tube
(a)
(c) e/m ratio is constant, independent of the gas filled and cathode material
(c)
2.
(d) none of these
(c) 300 nm
(d) 220 nm
(a)
n1
n2
(b)
n1
n2
(c) either, depending on how it is observed
8.
(d) neither The photoelectric current decreases if (a) the frequency of incident radiation decreases below threshold frequency (b) the exposure time is decreased (c) the intensity of the source of light is decreased
9.
(d) none of these
5.
h 2
(b) 400 nm
(b) a particle
4.
2( 2 1)
h 2
(a) 540 nm
What is an electron ? (a) a wave
3.
(b)
The hydrogen line spectra provides evidence for the (a) Heisenberg uncertainty principle (b) wave like property of light (c) diatomic nature of hydrogen (d) quantized nature of atomic energy states If an element emits photon in its ground state due to electronic transitition, then its (a) atomic number will increase (b) atomic number will remain unchanged (c) atomic number will decrease (d) mass number will decrease If an electron undergoes transition from n = 2 to n = 1 in Li2+ ion, the energy of photon radiated will be best given by (a) hv (b) hv1 + hv2 (c) hv1 + hv2 + hv3 (d) all of these Kinetic energy of an electron in hydrogen atom increases after transition from an orbit n1 to another orbit n2. Then
(d) particles forming them possess the same mass and same charge.
7.
h
The work function of a metal is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from the metal is approximately :
10.
The work function of a photoelectric material is 3.3 eV, its threshold frequency will be (a) 8 × 1014 Hz
(b) 5 × 1033 Hz
(c) 8 × 1010 Hz
(d) 4 × 1011 Hz
MARK YOUR RESPONSE
(c) it is not possible to predict (d) none of these
1.
2.
3.
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5.
6.
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9.
10.
IIT-JEE CHEMISTRY Challenger
24
11.
12.
13.
The ratio of the speeds of an electron in the first orbit of hydrogen atom to that in the 4th orbit of He+ is (a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1 Hydrogen atoms are excited to n = 4 state. In the spectrum of emitted radiation, number of lines in the ultraviolet and visible regions are respectively (a) 3; 1 (b) 1; 3 (c) 2; 3 (d) 3; 2 An electron travels with a velocity of x ms–1. For a proton to have the same de Broglie wavelength, the velocity will be approximately? (a)
14.
1840 x
(b)
x 1840
18.
19.
20.
de Broglie wavelengths Li and p have the ratio (assume 21.
9m p )
(a) 1 : 2 15.
(b) 1 : 4
(c) 1 : 1 (d) 1 : 3 3 When light of frequency 3.2 × 1016 Hz is used to irradiate a metal surface, the maximum kinetic energy of the emitted
3 photoelectron is of the energy of the irradiating photon. 4 What is the threshold frequency of the metal? (a)
2.4 10 25 Hz
(c) 1.6 1015 Hz 16.
17.
(b)
The ionization potential for the electron in the ground state of the hydrogen atom is 13.6 eV atom–1. What would be the ionization potential for the electron in the first excited state of Li2+? (a) 3.4 eV (b) 10.2 eV (c) 30.6 eV (d) 6.8 eV The energy of the electron in Be3+ ion depends on (a) the principal quantum number only (b) the principal and azimuthal quantum numbers only (c) the principal, azimuthal and magnetic quantum numbers only (d) the principal, azimuthal, magnetic and spin quantum numbers
MARK YOUR RESPONSE
22.
p
2.4 1016 Hz
(d) 8 1015 Hz
1 1 3 1 1 3 , , , , (b) 2 2 2 2 2 2 (c) +2, +1, 0, –1, –2 (d) +1, 0, –1 The average life of an excited state of hydrogen atom is of the order of 10–8 s. The number of revolutions made by an electron when it is in state n = 2 and before it suffers a transition to state n = 1, are (a) 2.28 × 106 (b) 22.8 × 106 6 (c) 8.23 × 10 (d) 2.82 × 106 What transition in the hydrogen spectrum would have the same wavelength as Balmer transition, n = 4 to n = 2 in He+ spectrum? (a) n = 3 to n = 1 (b) n = 3 to n = 2 (c) n = 4 to n = 1 (d) n = 2 to n = 1 The dissociation energy of H2 is 430.53 kJ mol–1. If hydrogen is dissociated by illumination with radiation of wavelength 253.7 nm, the fraction of the radiant energy which will be converted into kinetic energy is given by (a) 1.22% (b) 8.76% (c) 2.22% (d) 100% If a proton and -particle are accelerated through the same potential difference, the ratio of de Broglie wavelengths (a)
(c) 1840 x (d) x Li and a proton are accelerated by the same potential, their
mLi
What are the component values (in terms of h/2 ) of the orbital angular momentum along the Z-direction for a 2p electron?
23.
is
(a) 2
(b) 1
(c)
(d) 3
2 2
If the uncertainties in position and momentum are equal, the uncertainty in the velocity is (a)
(c) 24.
and
h
1 2m
(b) h
h 2
(d) none of these
The orbital angular momentum of 2p electron is (a)
3h
(b)
6h
(c) Zero
(d)
2
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
h 2
STRUCTURE OF ATOM 25.
In any subshell, the maximum number of electrons having same value of spin quantum number is (a) (c)
26.
27.
28.
l (l 1) 2l
2
30.
(b)
l
(d)
4l
(a)
(b)
(c)
(d)
31.
2 2
The orbital diagram in which both the Pauli’s exclusion principle and Hund’s rule are violated, is
The incorrect statement amongst the following is (a) Magnetic quantum number determines the zcomponent of the orbital angular momentum of the electron (b) In the applied magnetic field, d-orbitals can assume 5 different orientations w.r.t. the field (c) The wave function associated with half filled or completely filled sub-shells is spherically symmetrical (d) None of these Given the following statements regarding d-orbitals (1) It has 4 lobes and 2 angular nodes (2) It has 4 lobes and one angular node (3) The orbital angular momentum of electron in this orbital is
29.
25
h 6 2
32.
(b) equal to the circumference of the orbit (c) twice the circumference of the orbit (d) thrice the circumference of the orbit The de Broglie wavelength of an electron accelerated by an electric field of V volts is given by
1.23
(a)
m
1.23
(c) 33.
34.
35.
(4) It is non-degenerate The correct statement(s) is/are (a) only (1) (b) (1) and (3) (c) (2) and (3) (d) (1) and (4) If the shortest wavelength of the spectral line of H-atom in Lyman series is x, then the longest wavelength of the line in Balmer series of Li2+ is
The de Broglie wavelength of electron in first Bohr orbit is exactly equal to (a) half the circumference of the orbit
36.
V
nm
(b)
nm
(d)
nm
1.23 nm V
(a) 25 (b) 9 (c) 36 (d) none of these Which two orbitals are both located between the axes of coordinate system, and not along the axes? (a)
d xy , d
z2
(c)
d
, pz
x2 y2
(b)
d yz , px
(d) none of these
Which series of subshells is arranged in the order of increasing energy for multi-electron atoms? (a) 6s, 4 f, 5d, 6p (b) 4 f, 6s, 5d, 6p (c) 5d, 4 f, 6s, 6p (d) 4f, 5d, 6s, 6p The four quantum numbers that could identify the third 3p electron in sulphur are (a) n = 3, l = 0, m = +1, s =
1 2
5x 4
(b)
4x 5
(b) n = 2, l = 2, m = –1, s =
1 2
(c)
x 9
(d)
9x
(c) n = 3, l = 2, m = +1, s =
1 2
(d) n = 3, l = 1, m = –1, s =
1 2
MARK YOUR RESPONSE
h
The minimum number of orbitals possible for a shell containing g-subshell is
(a)
Which of the following electron transitions in a hydrogen atom will require the largest amount of energy? (a) from n = 1 to n = 2 (b) from n = 2 to n = 4 (c) from n = 10 to n = 1 (d) from n = 3 to n = 5
1.23
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
IIT-JEE CHEMISTRY Challenger
26
37.
The quantum numbers of four electrons are as follows : n l m s Electron 1
3
1
1
1 2
(ii) Electron 2
3
0
0
1 2
0
1 2
(i)
(iii) Electron 3
4
(iv) Electron 4
38.
39.
40.
41.
42.
3
0 2
2
2 a0
(b) 16a0
(c)
a0 4
(d)
45.
(a)
(rmax )3d
(rmax )3 p
(rmax )3s
(b)
(rmax )3d
(rmax )3s
(rmax )3 p
(c)
( rmax ) 3 s
The correct order of penetrating power of 3s, 3p, 3d electrons is (a) 3d > 3p > 3s
(b) 3s > 3p > 3d
(c) 3s > 3d > 3p
(d) 3d > 3s > 3p
Hund’s rule pertains to the distribution of electrons in (a) principal energy shell
(b) an orbital
(c) degenerate orbitals
(d) none of these
If there are three possible values
1 , 0, 2
46.
47.
(a) 2, 8, 8, 18
(b) 3, 12, 27, 27
(c) 3, 12, 12, 27
(d) 2, 8, 12, 27
for the
The number of radial nodes (nodal points) of 3s, 3p and 3d electrons are respectively (a) 0, 1, 2
(b) 2, 1, 0
(c) 1, 3, 5
(d) 3, 2, 0
When a certain photoelectric substance was irradiated with a light of frequency 4.0 × 1015 Hz, the photo electrons emitted had thrice the kinetic energy as did photo electrons emitted when the same substance was irradiated with light of frequency 2.0 × 1015 Hz. The threshold frequency of the photoelectric substance is 2.0 1015 Hz
(b) 1.0 1016 Hz
(c) 1.0 1015 Hz
(d) 1.5 1015 Hz
(a)
48.
49.
The wave number of last line of Lyman series of hydrogen spectrum is 109674 cm–1. The wave number of H line in Balmer series He+ is (a) 438696 cm–1
(b) 109674 cm–1
(c) 30465 cm–1
(d) 60930 cm–1
The energy of electron in first excited Bohr’s orbit of He+ is 13.6 eV atom–1. The energy of electron in the second excited orbit of Li2+ is (a) –30.6 eV atom–1 (c) –7.65 eV atom–1
( rmax ) 3d
(b) –13.6 eV atom–1 (d) –27.2 eV atom–1
(d) none of these
MARK YOUR RESPONSE
1 2
spin quantum numbers, the number of elements in first, second, 3rd and 4th periods of long form of periodic table will be respectively
8 a0
The radii of maximum probability for 3s, 3p and 3d electrons are in the order
( rmax )3 p
44.
1 2
The correct order of decreasing energies of these electrons is (a) Electron 2 > Electron 1 > Electron 3 > Electron 4 (b) Electron 4 > Electron 3 > Electron 1 > Electron 2 (c) Electron 4 > Electron 3 > Electron 2 > Electron 1 (d) Electron 1 > Electron 2 > Electron 4 > Electron 3 The principle that is based on electrons attempting to be as apart as possible is (a) Bohr’s theory (b) Heisenberg principle (c) Exclusion principle (d) Hund’s rule The atomic number of the element with maximum number of unpaired electrons is (a) 23 (b) 33 (c) 15 (d) 26 Which of the folloing orbitals is more close to the nucleus? (a) 5f (b) 6d (c) 7s (d) 7p If the radius of first orbit of H-atom is a0, then de Broglie wavelength of electron in 4th orbit is (a)
43.
37.
38.
39.
40.
41.
42.
43.
44.
45.
46.
47.
48.
49.
STRUCTURE OF ATOM 50.
51.
27
Given the following ions (1) Cr3+ (2) Fe2+ (3) Ni2+ (4) Cu2+ (atomic number : Cr = 24, Fe = 26, Ni = 28, Cu = 29) The correct sequence of the increasing order of number of unpaired electrons in these ions is (a) 3, 4, 2, 1 (b) 1, 2, 3, 4 (c) 4, 3, 1, 2 (d) 4, 2, 3, 1 The magnetic moment of Mx+ (atomic number of M = 25) is
57.
(a) [Ne] (b) [Ne] (c) [Ne]
15 B.M. The number of unpaired electrons and the value
52.
53.
of x respectively are (a) 4, 3 (b) 3, 4 (c) 3, 2 (d) 5, 2 The number of d-electrons in Fe2+ (Z = 26) is not equal to that of (a) s-electrons in Mg (b) p-electrons in Ne (c) d-electrons in Fe (d) p-electrons in Cl The ratio of magnetic moments of Fe(III) and Co (II) is (a)
54.
55.
56.
(b)
3: 7
(a) 1 : 5 : 4
(b)
(c) 1 :2:6
(d) 1 : 3 : 21
(c)
2s
2p
2s
2p
MARK YOUR RESPONSE
(b) (d)
58.
2s
2p
2s
2p
3s
3p
3s
3p
3s
3p
3s
3p
The total number of orbitals in the principal shell of He+ that has energy equal to
59.
hcR , is (R = Rydberg constant) 4
(a) 4 (b) 16 (c) 9 (d) none of these Which of the following sets of three quantum numbers
(n, l and m) that does not give the permissible solution
60.
4 : 5 :1
Which of the following is not correct about the product of velocity of electron in Bohr’s orbit and the principal quantum number in a hydrogenic atom or ion? (a) It is independent of the principal quantum number (b) It is independent of the energy of electron (c) It has no dependence of nuclear charge (d) It varies proportionately with the nuclear charge Which of the following orbital diagram violates the Pauli’s exclusion principle? (a)
(d) [Ne]
7: 3
(c) 7 : 3 (d) 3 : 7 The magnetic moments of Cu (Z = 29), Ti (Z = 22), and Cr (Z = 24) are in the ratio of
The orbital diagram in which the Aufbau principle is violated, is
61.
of Schrodinger wave equation? (a) 3, 2, 2 (b) 4, 2, 0 (c) 4, 3, –3 (d) 4, 2, 3 If the subsidiary quantum number of a sub-energy level is 4, the maximum and minimum values of the spin multiplicities are (a) 9, 1 (b) 10, 1 (c) 10, 2 (d) 4, –4 The correct set of quantum numbers for the unpaired electron of scandium (Z = 21) is (a) 4, 2, 2,
1 2
(b) 3, 2, –2,
(c) 3, 1, 1,
1 2
(d) 3, 2, 3,
1 2 1 2
62.
The orbital of next higher energy than that of n p orbital is (a) nd (b) (n + 1)p (c) (n + 1) d (d) (n + 1) s
63.
The magnetic moment of a particular ion is 2 6 B.M. The ion is (a) Mn 2+ (b) Fe3+ 2+ (c) Co (d) Co3+
50.
51.
52.
40.
53.
54.
55.
56.
57.
58.
59.
60.
61.
62.
63.
IIT-JEE CHEMISTRY Challenger
28
64.
The electronic configuration of
3d
Mn2+
is
67.
4s
(a)
(a) [Ar]
3d
4s
3d
4s
3d
4s
68.
(c) [Ar]
(d) [Ar]
(b) 1
(c) 2
(d) 3
2 2
Which of the following graphs represents the radial charge density of 3d electron?
69.
70.
71.
4 rR
(a)
r
4 rR
2 2
72.
(b)
73.
r
(d) pz, d
8 RH 9 In order to have a wavelength of 1Å associated with an electron, the accelerating electric potential must be (a) 10 V (b) 150 V (c) 15 V (d) 1 V A principal shell having the highest energy subshell to be ‘g’ can accomodate electrons to a maximum of (a) 18 (b) 32 (c) 25 (d) 50 A 5 ‘g’ orbital has (a) 0 angular and radial nodes both (b) 0 radial node and 2 angular nodes (c) 4 radial nodes and 4 angular nodes (d) 0 radial node and 4 angular nodes A 3 p-orbital has (a) 2 non-spherical nodes (b) 2 spherical nodes (c) 1 spherical and 1 non-spherical node (d) 1 spherical and 2 non-spherical nodes If n and l are the principal and azimuthal quantum numbers, then maximum number of electrons that a principal energy level can accomodate, is (d)
2 2
l n
2 (2l 1)
(a)
l 0 l n 1
l n 1
2(2l 1)
(c)
2 2
(d)
r
MARK YOUR RESPONSE
2 (2l 1)
(d) l 1
l 0
74.
2(2l 1)
(b)
l 1
r
The energy of an electron in first Bohr orbit of H-atom is – 13.6 eV. The possible energy value of electron in the excited state of Li2+ is (a) – 122.4 eV (b) 30.6 eV (c) – 30.6 eV (d) 13.6 eV
64.
65.
66.
67.
68.
69.
70.
71.
72.
73.
74.
x2 y 2
The wave number of electromagnetic radiation emitted during the transition in between two energy levels of Li2+ ion whose principal quantum number sum is 4 and difference is 2, is (a) 3.5 RH (b) 4 RH
l n 1
4 rR
(c)
(b) dxy, pz
z2
(c) 8 RH
4 rR
66.
The number of concentric spherical surfaces for 3s orbital at which the probability of finding the electron is zero, are (a) 0
d xy , d
(c) dyz, px
(b) [Ar]
65.
Which two orbitals are both located along the axes, and not between the axes?
STRUCTURE OF ATOM 75.
29
The kinetic energy of an electron in an orbit of radius r of
Li (a)
2
ion is (e = electronic charge)
e2 r
(b)
e2 3r
82.
2
3e 3e 2 (d) 2r r Magnetic field produced by electrons in atoms and molecules is due to their (a) orbital motion only (b) spin motion only (c) spin and orbital motion both (d) neither spin nor orbital motion When an electron of H-atom jumps from a higher to lower energy state, then (a) its potential energy increases (b) its kinetic energy increases (c) its angular momentum remains unchanged (d) wavelength of de Broglie wave associated with the electron increases An energy of 24.6 eV is required to remove one of the electrons from a helium atom. The energy required to remove both the electrons from helium atom is (a) 38.2 eV (b) 49.2 eV (c) 51.8 eV (d) 79.0 eV As per Bohr model the potential energy of electron in the
(c) 76.
77.
78.
79.
80.
81.
ground state of Li 2 is (a) – 27 eV (b) – 122.4 eV (c) – 244.8 eV (d) 244.8 eV Electronic configuration of a species (atom or ion) depends upon (a) total number of electrons present only (b) nuclear charge only (c) total number of electrons as well as nuclear charge (d) total number of electrons as well as number of neutrons The electron in H-atom like species makes a transition from an excited state to the ground state. Which of the following statements is true? (a) Its kinetic energy increases but potential and total energy decreases (b) Its kinetic energy decreases whereas the potential and total energy remains the same
MARK YOUR RESPONSE
83.
(c) Its kinetic and total energy decreases and potential energy increases (d) Its kinetic, potential and total energy decreases If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of Li2+ is (a) 3.4 eV (b) 13.6 eV (c) 30.6 eV (d) 122.4 eV The shortest wavelengths of radiation emitted when an electron from infinity falls into the ionized H-atom ( RH
1.097 107 m
(a) 91 nm 84.
85.
87.
88.
) (b) 192 nm
(c) 406 nm (d) 1.9 10 8 nm According to Einstein’s Photoelectric equation, the plot of kinetic energy of photoelectron from a metal surface versus reciprocal of wavelength of incident radiation is a straight line. The slope of the line (a) depends on the intensity of radiation and metal used (b) depends on the nature of the metal used (c) depends on the intensity of the radaition used (d) is the same for all metals and independent of the intensity of radiation Photoelectric work function of a metal is 1 eV. If a light of wavelength 300 Å falls on the metal, the velocity of the photoelectron ejected will be (a) 10 ms
86.
1
1
(b) 10 3 ms
1
(c) 10 4 ms 1 (d) 10 6 ms 1 The work function of a metal is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this metal is approximately (a) 540 nm (b) 400 nm (c) 310 nm (d) 220 nm The binding energy per nucleon of deutron (H2 is 1.1 mev and that of He nucleus is 7 mev. When two deutron nuclei fuse to form helium nucleus, the energy released is (a) 13.9 mev (b) 23.6 mev (c) 29.6 mev (d) 19.2 mev The electronic configuration 1s 2 2 s 2 2 p 5 3s1 describes which one of the following (a) The ground state of neon (b) The ground state of F anion (c) The ground state of oxide anion (d) The excited state of oxide anion
75.
76.
77.
78.
79.
80.
81.
82.
83.
84.
85.
86.
87.
88.
IIT-JEE CHEMISTRY Challenger
30
89.
Which of the following graphs depicts correctly the variation of photoelectric current, i, with the intensity, I, of incident radiation? (a)
93.
(b) 94.
i
i
I (c)
I 95.
(d)
i
Ii I
I 90.
91.
If it were possible for hydrogen atom to exist with a positron as the extra nuclear particle, then the energy of positron in first excited state will be (a) 13.6 eV (b) 3.4 eV (c) – 3.4 ev (d) 6.8 eV What will be the number of spectral lines (N) observed if an electron undergoes transition from n2 excited level to n1 excited level in an atom of hydrogen ? (a)
N
(b)
N
n1 )(n2 2
n1 1)
(n1 n2 )(n2 2
n1 1)
97.
98.
n1 )(n1 n2 1) 2 (d) 2(n1 – n2)(n2 + n1 – 1)2 If the shortest wave-length of a spectral line in Lyman series for He+ is ‘x’, then what will be the largest wave length of a spectral line in Balmer series of Li2+ (c)
92.
(n2
96.
The correct relationship between wave numbers of spectral lines, in Balmer series of hydrogen atom having the least and highest wave-lengths, is given by (a) 5 : 9 (b) 9 : 5 (c) 5 : 27 (d) 27 : 5 If we apply potential difference so that an electron is accelerated continuously in a vaccum tube such that a decrease of 10% occurs in its de Broglie wave length. In such a case the change observed in kinetic energy of electron will be approximately (a) a decrease of 11% (b) an increase of 11.1% (c) an increase of 10% (d) an increase of 23.4% The approximate wave length of matter wave associated with an electron, that is accelerated by applying 100 V of potential difference in a discharge tube, will be (a) 123 pm (b) 12.3 pm (c) 1.23 pm (d) 0.123 pm Which of the following gives the correct ratio between time periods of an electron for each revolution in second and third Bohr’s orbits ? (a) 4 : 9 (b) 9 : 4 (c) 8 : 27 (d) 27 : 8 Which of the following is the correct sequence for the number of nodal planes, number of nodes and number of peaks in the radial probability curve of a ‘5d’ orbital ? (a) 2, 3, 2 (b) 2, 2, 3 (c) 4, 3, 2 (d) 5, 2, 3 Energy levels, A, B, C, of a certain atom correspond to increasing values of energy i.e., EA < EB < EC. If 1, 2, 3 are the wave lengths of radiations corresponding to the transition from C to B, B to A and C to A respectively, which of the following statements is correct ?
(a)
(c)
N
(n2
4x 5
1
(d)
B 2
3
A
(b) 9 x
5x 4
MARK YOUR RESPONSE
C
16 x 5
(a)
3=
1+ 2
(b)
3
(c)
1+
2+
(d)
2 3
3=0
89.
90.
91.
92.
93.
94.
95.
96.
97.
98.
1 2 1
2
2 1
2 2
31
STRUCTURE OF ATOM 99.
The given diagram indicates the energy levels of certain atoms. When the system moves from 2E level to E a photon of wave length is emitted. The wave-length of photon 4E produced during its transition from level to E is 3
102.
If E is the kinetic energy of the particle then which of the following expressions is correct for de Broglie wave length of the particle ? (a)
=
h 2mE
(c)
=
h
(b)
=
h 2mE
(d)
=
2mE h
2E 4 3E E
?
103.
0 (a)
(c) 100.
101.
(b)
3 4 3
3 3
104.
(d) 3
The transition from state n = 4 to n = 3 in a hydrogen like atom results in ultraviolet radiations. Infra-red radiation will be obtained in which of the following transitions ? (a) n = 2 to n = 1 (b) n = 3 to n = 2 (c) n = 4 to n = 2 (d) n = 5 to n = 4 Radial part of the wave function depends upon quantum numbers (a) n and s (b) l and m (c) l and s (d) n and l
MARK YOUR RESPONSE
99.
100.
104.
105.
PASSAGE-1 Dual Nature of Matter : Light and other electromagnetic radiations have dual nature i.e., wave and particle nature. Interference and diffraction phenomenon show the wave nature of light whereas the photoelectric effect and black body radiation phenomenon support the particle nature of light. Louis de Broglie (1924) suggested on the same grounds that all moving material particles (microscopic as well macroscopic) also have the dual nature i.e., wave and particle nature. However, the wave associated with matter, known as matter wave, has characteristics different from the electromagnetic wave. The two characters of matter are related by de Broglie equation as
=
h h = mv p
......
(h = Planck's constant)....(1)
105.
2mE
Angular part of the wave function depends upon quantum numbers (a) n and l (b) l and m (c) n and s (d) l and s If two particles are associated with same kinetic energy, then the de Broglie’s wave-length ( ) of these particles is (a) directly proportional to the velocity (b) inversely proportional to the velocity (c) independent of mass and velocity (d) can not be predicted An electron in the ground state of hydrogen was excited to a higher energy level using monochromatic radiations of wave length ( ) 975 Å. The longest wave length that appears in the resulting spectrum is due to transition from: (a)
n4
n1
(b)
n4
n3
(c)
n5
n4
(d)
n5
n1
101.
102.
103.
where denotes the wavelength associated with the moving matter object of mass m and velocity v, and p denotes the momentum of the object. Davisson and Germer determined the wavelength of matter wave associated with electrons (accelerated by an exactly known electric potential) by studying their diffraction patterns from nickel crystal and using the relation: 2d sin = n .....(2) where d denotes the spacing of lattice planes in nickel crystal, the grazing angle of incidence on the planes and n (1, 2, 3 etc.) the order of reflection. The experimentally determined value of from Eq. (2) is found to be in excellent agreement with the value of wavelength calculated using de Broglie equation as : =
h = mv
h 2meV
IIT-JEE CHEMISTRY Challenger
32
where e is the electronic charge (coulombs) and V (volts) is the accelerating potential. 1.
The momentum of hydrogen atom is given by (a)
p
(a)
mc2
h 10
4
m
me
me 10 h
2
(b)
m
h 10 me
2
m
me 10 4 m h
(d)
If is the wavelength associated with the electron in the second circular orbit of H-atom, the radius of the orbit is given by (a)
(b)
2
(c) 4.
p
p
(c) 3.
(b)
hv
mv (d) Can't be predicted h The wavelength of matter wave associated with an electron passing through an electric potential of 5000V is given by (e = electronic charge in coulombs and m = mass of electron in Kg.)
(c)
2.
hv c
2
If the Planck's constant is 6.6 × 10–34 Js, the de Broglie wavelength of a particle of mass 3.3 × 10 –27 kg and velocity 200 ms–1 will be (a) 2 Å (b) 5 Å (c) 20 Å (d) 10 Å PASSAGE-2
Erwin Schrodinger developed an atomic model which is based on the particle and wave nature of the electron. This model describes the electron as a three dimensional stationary wave in the electronic field of positively charged nucleus. Schrodinger derived an equation which describes wave motion of an electron. The equation, widely known as Schrodinger equation, is
x
2 2
MARK YOUR RESPONSE
y
2 2
z
1.
8 2
2
h
m
2
the electron. The wave function may be reagarded as the amplitude function in terms of coordinates x, y and z. It may have positive or negative values depending upon the values of coordinates. As such has no significance but its square ( 2 ) gives the probability of finding an electron within a small element of space. The narrow region of space in which there is maximum probability of finding the electron is called orbital. The solution of the above equation requires three sets of numbers to be introduced, called quantum numbers. These are denoted by n, l and m and called principal, azimuthal and magnetic quantum numbers respectively. n assumes integral values (1, 2, 3, 4 etc.), l assumes values 0 to n – 1 (all integral) and m assumes –l to +l through zero (all integral). The principal quantum number describes the size and the energy of the orbital, l describes the shape and m the orientation of the orbital in the applied magnetic field. The hyperfine splitting of spectral lines of H-spectrum under the electric field necessitates the introduction of fourth quantum number, the spin quantum number(s). Spin quantum number can take up two values
(d)
2
where x, y, z are cartesian coordinates of the electron m = mass of electron; E = total energy of the electron; V = potential energy of the electron; h = Planck's constant and (psi) = wave function of
(E V )
2.
0
1 and 2
1 which 2
describe the spin angular momentum of the electron in clockwise and anticlockwise directions. The four quantum numbers address completely a particular electron. No two electrons in an atom can have the same set of the four quantum numbers, if three numbers have the same values for the two electrons, the fourth (say s) will differ (Pauli's exclusion principle). Orbitals are designated by specific values of n and l. Their energies are determined by the value of n + l in multielectron atom, larger is the value of n + l, the greater is the energy of the orbital. For Hatom and like species (He+, Li2+ etc.), only the principal quantum number determines the energy of the orbital. Various orbitals are filled with electrons in the increasing order of their energy (Aufbau principle). Degenerate orbitals (orbitals of equal energy) of a particular subshell, designated by particular set of n + l values, are filled singly first with parallel spins; pairing of electrons (with opposite spins) starts when all the degenerate orbitals are singly occupied. (Hund's rule)
3.
4.
STRUCTURE OF ATOM 5.
6.
7.
8.
Which of the following quantum numbers is not the direct consequence of wave mechanical model of an atom? (a) Principal quantum number (b) Azimuthal quantum number (c) Magnetic quantum number (d) Spin quantum number Which of the following is independent of all the three quantum numbers n, l and m in a multielectron atom? (a) Energy of electron (b) Orbital angular momentum (c) Orientation of orbital (d) Spin angular momentum Which set of orbitals in Li2+ represents degeneracy? (a) 2s, 3s, 4s (b) 2s, 3p, 3d (c) 3s, 3p, 3d (d) None of these The orbital diagram in which 'Aufbau' principle is violated, is (a) (c)
9.
33
2s
2p
2s
2p
(b) (d)
2s
2p
2s
2p
(III)
The magnetic quantum number (m) determines the direction of rotation of the electron. The mathematical functions for atomic orbitals may be written as a product of two factors. (i) The radial wave function describes the behaviour of an electron as a function of distance from the nucleus. (ii) The angular wave function shows how it varies with the direction in space. Angular wave functions do not depend on n and are features of s-, p-, d- ..... orbitals. Diagrammatic representations of angular functions for s-, p-, d- orbitals are shown in figure below
Which of the following statements is correct for an electron that has n = 4 and m = –2? (a) The electron is in d-orbital (b) The electron is in a p-orbital (c) The electron is in fourth principal electronic shell (d) The electron must have the spin quantum number to be
1 2 Fig. Shapes of s and p orbitals Mathematically, they are essentially Polar diagrams showing how the angular wave function depends on the polar angles and . They can be regarded as boundry surfaces enclosing the regions of space where the electron is most likely to be found. An s-orbital represented by a sphere, the probability is same in all directions in space. Each p-orbital has two lobes, with positive and negative values of the wave function on either side of the nucleus, separated by a nodal plane where the wave function is zero. The five d-orbitals each have two nodal planes, representing two positive and two negative regions of wave functions. The f-orbitals each have three nodal planes.
PASSAGE-3 The atomic orbitals of hydrogen are labelled by quantum numbers. Three integers are required for a complete specification. (I) The principal quantum number (n) determines how far from the nucleus the electron is most likely to be found. (II) The angular momentum (or azimuthal) quantum number (l) determines the total angular momentum of the electron about the nucleus. mvr
l (l 1).
MARK YOUR RESPONSE
h , where mvr = angular momentum. 2
5.
6.
7.
8.
9.
IIT-JEE CHEMISTRY Challenger
34
10.
11.
The incorrect statement amongst the following is (a) Magnetic quantum numbers determine the z-component of orbital angular momentum of the electron
3p
(b) Under the influence of applied magnetic field, the dorbitals can assume 5 different orientations
2p
(c) The wave function associated with half-filled or completely filled sub-shells is spherically symmetrical
1s
(d) none of these
1s
The de Broglie's wave length ( ) of an electron accelerated by an electric field of V volts is given by (a)
(c) 12.
3s
1.23 10 9 m 1.23 10 9 V
m
m
(b)
(d)
1.23 10
1.23 10 9
500
1000
m
m
In 5g-orbitals, we have (a) angular nodes = 0, radial nodes = 0 (b) angular nodes = 2, radial nodes = 0
Radial distributions determine the energy of an electron in an atom. The subsidiary maxima at smaller distances are not significant in hydrogen, but are useful in understanding the energies in many electron atoms. The energies of atomic orbitals in hydrogen atom are given by the formula
(c) angular nodes = 4, radial nodes = 4
Radial wave function depends on n and l but not on m. Thus each of the three p-orbitals have the same radial form. The wave functions may have positive or negative regions but their radial probability distributions (figure below) show the following features. Radial distributions may have several peaks, the number being equal to (n – 1). The outermost peak is by far the largest, showing where the electron is most likely to be found. The distance of this peak from the nucleus is a measure of the radius of the orbital and is roughly proportional to n2 (although it slightly depends on l also)
All orbitals with finite n represent bound electrons with lower energy. Energies of individual atoms or molecules are expressed in electron volts (eV) equal to about 1.602 × 10–19 J. For many electron atoms The orbital sizes and energies depend on the atomic number ‘Z’ Average radius of an orbital
10.
11.
n2 a0 Z
Where a0 = Bohr’s radius (59 pm) i.e., radius of 1s orbital of hydrogen atom Also En
MARK YOUR RESPONSE
n2
This shows that the energy depends only on the principal quantum number, n.
PASSAGE-4
•
R
En
(d) angular nodes = 4, radial nodes = 0
•
1500
radius (pm)
9
h
E
0
12.
Z 2R n2
STRUCTURE OF ATOM 13.
35
The total number of sub-shells in the principal shell of He+ having energy equal to
hcR , is 4
15.
(R = Rydberg constant)
14.
2.
3.
(c) 10, 1
(d) 10, 2
The number of d-electrons in Fe2+ is not equal to that of (a) d-electrons in Iron
(b) 2
(b) p-electrons in Neon
(c) 3
(d) 4
(c) p-electrons in Chlorine
If the subsidiary quantum number of a sub-energy level is 4, the maximum and minimum values of the spin multiplicities are given by
14.
13.
:
The kinetic energy of photoelectron ejected increases with the frequency of incident light.
Statement-2
:
Increase in intensity of incident light increases the photoelectric current.
Statement-1
:
6.
:
Threshold frequency characteristic to a metal.
Statement-1
:
Spin quantum number can have two
:
+ and – signs signify the positive and negative wave functions.
Statement-1
:
Electromagnetic waves consist of oscillating electric and magnetic fields.
6.
7.
Matter waves are radiated into the space.
Statement-1
:
The transition of electrons
Statement-2
:
is
Statement-2
2.
:
n2 in H-atom will emit
radiation of higher frequency than n4 n3. Principal shells n2 and n3 have lower energy than n4 . 7.
1 1 and – . 2 2
1.
Statement-2
n3
Threshold frequency is the maximum frequency required for the ejection of electron from the metal surface.
Statement-2
MARK YOUR RESPONSE
(d) s-electrons in Magnesium
15.
Statement-1
values, +
45.
(b) 9, 1
(a) 1
MARK YOUR RESPONSE
1.
(a) 4, –4
8.
Statement-1
:
An orbital designated by n = 3, l = 1 has double dumb-bell shape.
Statement-2
:
It belongs to p-subshell.
Statement-1
:
Electromagnetic radiation will be emitted for the transition
2 pz
3.
4.
2 px .
5.
IIT-JEE CHEMISTRY Challenger
36
Statement-2
:
No transition of electron takes
9.
Statement-1
:
When electron of He
10.
11.
12.
is
Statement-2
:
He 2
Statement-1
:
Potassium chloride imparts purple colour to the flame.
Statement-2
:
Potassium has low ionization energy.
Statement-1
:
It is not possible to find an electron present in the nucleus.
Statement-2
:
Velocity of electron-wave is less as compared to velocity of light.
Statement-1
:
Angular momentum of an electron in any orbit is given by angular
MARK YOUR RESPONSE
principal quantum number.
is raised
from 1-s to infinity, He 2 formed.
13.
9.
13.
14.
14.
Which of the following sets of quantum numbers is/are not allowed? (a) n = 3, l = 2, m = –1 (c) n = 3, l = 0, m = 1
2.
:
The principal quantum number, n, can have any integral value.
Statement-1
:
Sommerfield, in 1915, proposed that electrons revolve around the nucleus in elliptical path.
Statement-2
:
Sommerfield extension is helpful in explaining the different intensities as well as the splitting of spectral lines in the electric and magnetic fields.
Statement-1
:
The presence of electrons in the inner orbits reduces the attraction between the nucleus and the outer electrons.
Statement-2
:
For any electrons, screening constant, may be calculated by Slater’s rule.
10.
3. 1.
Statement-2
has still one electron.
8.
(b) n = 2, l = 3, m = –1 (d) n = 6, l = 2, m = –1
dxy orbital has four lobes between x- and y-axes. The wave functions of two lobes are positive and those of other two are negative. The positive wave function signifies that
4.
(a) both x and y are positive
11.
0.5
(c) either x or y is negative (c) 1.5
(d) none of these
1.
2.
3.
12.
Which of the following statements is/are true about the quantum numbers n, l, m and s? (a) l gives an idea of the shape of the orbital (b) m gives the information of the energy of electron in a given orbital (c) n gives the idea of the size of the orbital (d) s gives the direction of spin of electron in an orbital The angular momentum of electron can have the value(s) (a)
(b) both x and y are negative
MARK YOUR RESPONSE
n.h , where n is the 2
momentum
place between 2 pz and 2 px orbitals.
h
(b)
h 2
(d)
4.
h
2
h
STRUCTURE OF ATOM 5.
37
The energy of an electron in the first Bohr orbit of H-atom is –13.6 eV. Then, which of the following statement(s) is/ are correct for He+?
10.
(a) The energy of electron in second Bohr orbit is –13.6 eV
11.
(d) The speed of electron in the second orbit is 2.19 × 106 ms–1 6.
13.6Z 2
En
(b) The kinetic energy of electron in the first orbit is 54.46 eV (c) The kinetic energy of electron in the second orbit is 13.6 eV
For which of the following species, the expression for the energy of electron in nth orbit
n2
9.
has the validity? (b) Li2+
(c) Deuterium
(d) Tritium
According to Bohr’s atomic theory, which of the following relations is/are correct? (a) Kinetic energy of electron
Which of the following statement(s) is/are correct?
Z2 n2
(a) Fe3+ and Mn2+ have the same paramagnetic character
(b) The product of velocity of electron and the principal quantum number Z2
(b) Cu2Cl2 and CuCl2 are coloured
(c) Frequency of revolution of the electron in an orbit
Z2 n3
(d) The magnetic moment of both Fe 2+ and Co3+ is 2 6 B.M.
8.
1
(a) He2+
(c) MnO4– has purple colour and hence has 3d unpaired electrons
7.
eV atom
M+ ion is isoelectronic with argon and has Z + 1 neutrons (Z = atomic number of M+). The mass number of the element M will therefore be (a) 38
(b) 39
(c) 40
(d) 37
(d) Coulombic force of attraction on the electron 12.
3s
3p
3s
3p
3s
3p
3s
3p
(a) [Ne]
(a) Zn2+
(b) Cu+
(c)
(d)
(b) [Ne]
O 22
(c) [Ne]
For the radial probability distribution curves
(4 r 2 .Rn, l versus r graph ) , which of the following is/are
(d) [Ne]
correct? 13.
(a) The number of maxima is n l
If mp is the mass of proton, mn that of a neutron, M1 that of 10 Ne
(b) The number of nodal points is n l 1
20
nucleus and M2 that of 20 Ca 40 nucleus, then
which of the following relations is/are not true?
(c) The radius of maximum charge density (rmax) increases in the order; 3s < 3p < 3d
(a) M2 > 2M1
(b)
M2
20(m p
(d) The number of angular nodes is l
(c)
M2
(d)
M2
2 M1
MARK YOUR RESPONSE
n4
Ground state electronic configuration of P atom can be represented as
Which of the following ions is/are paramagnetic?
He 2
Z3
2 M1
5.
6.
7.
8.
10.
11.
12.
13.
9.
mn )
IIT-JEE CHEMISTRY Challenger
38
14.
17.
Which is correct in case of p-orbitals? (a) They are spherically symmetrical
The possible path(s) of cathode rays that are ejecting from cathode surface can be shown as
(b) They have strong directional character (a)
(c) They are three fold degenerate
–
(b)
–
(d) Their charge density along x, y and z-axes are zero 15.
In the equation
1
R
1
1
22
n2
(a) ideal gas constant (c) Rydberg constant 16.
, R is known as
(b) Boltzman constant (d) Balmer constant
18.
A.
Eleven p electrons
B.
one electron with ml = 2
C.
Two electrons with n = 3, l = 2
(a) Cl, Ti, Sc
(b) Cl, Sc, Ti
(c) Cl, V, Cr
(d) Cl, Ti, V
15.
14.
According to Bohr’s model of H-atom if Vn denotes the potential energy of electron, Kn the kinetic energy, En the total energy and rn the radius of nth orbit, then match the following : Column 1
Vn (A) K n
Column II
?
p.
1. MARK YOUR RESPONSE
0
–
(d)
–
According to Bohr’s theory (a) when a required amount of energy is supplied to an electron in an atom it jumps from lower orbit to higher orbit and remains there
Identify the elements of lowest atomic numbers which have the characteristics listed as below
MARK YOUR RESPONSE
1.
(c)
(b) when a required amount of energy is supplied to an electron in an atom it jumps from lower orbit to higher orbit and remains there for very short interval of time and returns back to lower orbit, radiating energy (c) the angular momentum of an electron is proportional to its quantum number, n (d) the angular momentum of an electron is independent of its quantum number, n
16.
17.
(B) (C)
18.
rn
E nx , x = ?
q.
–2
1
Z y, y
r.
–1
s.
1
rn
?
(D) Angular momentum in lowest orbital
STRUCTURE OF ATOM 2.
39
Electron of charge e (coulombs) and mass me(kg) is accelerated by a potential of 50V. Match the following: Column 1
Column II
(A) Velocity of
p.
h 10 eme
electron (ms–1) (B) K.E. of electron (J)
q.
10 eme
(C) Momentum of electron r.
50 e
(D) Wavelength of wave
10 e / me
s.
5.
associated with e– 3.
Match the electronic configurations listed in Column II with the descriptions listed in Column I : Column 1
(A) Shortest wavelength
Column II
(A) Violation of
4x 5
q.
4x 9
r.
x 9
s.
4x 27
p.
(B) Violation of Pauli’s
(B) Longest wavelength in Lyman series
q.
(C) Shortest wavelength
exclusion principle (C) Violation of
in Balmer series
r.
Hund’s rule
(D) Longest wavelength
(D) Violation of both
s.
in Balmer series 6.
Pauli’s and Hund’s rules Match the following for singly ionized helium atom if total energy of electron in first orbit in H-atom is –13.6 eV atom–1
MARK YOUR RESPONSE
p.
in Lyman series
Aufbau’s rule
4.
Column 1 Column II (A) P.E. of electron in p. –54.4 eV atom–1 ground state (B) K.E of electron in q. 13.6 eV atom–1 ground state (C) Total energy of r. –108.8 eV atom–1 electron in ground state (D) Ionization energy of s. 54.4 eV atom–1 + He in lowest excited state If the shortest wavelength of spectral line of H-atom in Lyman series is x, then match the following for Li2+ Column I Column II
2.
3.
5.
6.
Column - I (A) Lyman series (B) Balmer series (C) Pfund series (D) Bracket series
Column - II p. n2 = 2 q. n2 = 3 r. n2 = 5 s. n2 = 6
4.
IIT-JEE CHEMISTRY Challenger
40
1.
One mole of He+ ions are excited, spectral analysis showed that 50% ions are in 3rd level, 25% in second level and remaining in ground state. Ionization energy of H-atom is 13.6 eV atom–1. What is the total energy released (in terms of 103 kJ) when all the excited ions return to ground state ?
2.
Calculate the ratio of wavelength of first line of Balmer series of H-atom to the wavelength of first line of Lyman series of 10 times ionized sodium atom.
3.
To stop the flow of photoelectrons produced by electromagnetic radiation incident on a certain metal, a
1. MARK YOUR RESPONSE
2.
3.
negative potential of 300 V is required. If the photoelectric threshold of metal is 1500 Å, what is the frequency of the incident radiation (in terms of 1016 Hz)? 4.
Find the wavelength of first line of He + ion spectral series whose interval between extreme lines is 2.7451 × 104 cm–1.
5.
Ultraviolet light of wavelength 800 Å and 700 Å when allowed by fall on hydrogen atoms in their ground state is found to eject electrons with kinetic energy 1.8 eV and 4.0 eV respectively. Compute the value of Planck’s constant (in terms of 10–34 Js)?
4.
5.
STRUCTURE OF ATOM
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.
(b) (c) (c) (c) (a) (a) (d) (c) (a) (b) (b) (d) (b) (d) (d)
16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.
(d) (b) (c)
1 2 3
1 2
(b) (d)
1. 2. 3.
(b,c) (a,b) (a,c,d)
1. 3. 5.
41
(c) (a) (d) (c) (d) (b) (c) (c) (d) (b) (d) (d) (b) (b) (a)
31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45.
(d) (d) (d)
4 5 6
(c) (d)
3 4
4. 5. 6.
5 6
(a,c,d) (a,b,c,d) (a,d)
(b) (c) (a) (d) (a) (d) (b) (d) (d) (c) (d) (c) (c) (c) (c)
46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60.
7 8 9
(c) (b) (c)
(c) (b)
7 8
7. 8. 9.
(b) (c) (a,b,d)
A - q; B - r; C - s; D - p A - s; B - r; C - q; D - p A - r; B - s; C - q; D - p
1
3.31
2
653
(b) (c) (c) (b) (c) (b) (d) (b) (a) (c) (c) (d) (b) (d) (d)
61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75.
10. 11. 12.
7.45
4
76. 77. 78. 79. 80. 81. 82. 83. 84. 85. 86. 87. 88. 89. 90.
13 14 15
(c) (a)
9 10
2. 4. 6.
3
(d) (c) (d)
10 11 12
(d) (d)
(b) (d) (d) (c) (c) (a) (d) (c) (b) (d) (d) (c) (c) (c) (d)
(b,c,d) (a,c,d) (a,d)
11 12
13. 14. 15.
(c) (b) (d) (c) (c) (a) (c) (a) (d) (d) (c) (b) (d) (c) (b)
91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
(a) (d) (b) (d) (a) (c) (a) (b) (d) (d) (a) (c) (b) (a) (b)
13 14
(b) (b)
(d) (d) (c)
(b) (b)
(a,c) (b,c) (c)
16. 17 18
(b) (a,b,c) (b,c)
A - s; B - r; C - q; D - p A - r; B - s; C - p; D - q A - p, q, r, s; B - q, r, s; C - s; D - r, s
4689
5
6.57
IIT-JEE CHEMISTRY Challenger
42
3. 4.
(c) Photoelectric current is directly propotional to the intensity of incident radiation having threshold frequency. (c) Work function = 4.0 eV = 4.0 × 1.6 × 10–19 J
hc
o
8.
9. 10.
11.
3.3 1.6 10
19
34
Js
6.6 10
J
9
14
8 10 s
16.
h me x
(b)
p
He
Z
h m pv
He
nH
Z 2.19 106 ms n
1 4 2 1
h 1840mev
h 2eVm p
;
2 3eV
3.2 1016 4
o
4
E
13.6 Z 2
1
17. 18.
8 1015 Hz
1
1
n12
n22
eV atom –1
2 and n2
1
= 30.6 eV 2 2 (a) Be3+ is hydrogenic ion, i.e. consists of one extranuclear electron. (d) The component values of orbital angular momentum 2
h 2
Z ms n
2.19 106
(c) Velocity of electron, vn
1
(Z = 1
for H) The distance travelled by the electron in second Bohr’s orbit in 10
8
s
2.19 106 1.0 10 8 m 1.095 10 2 The circumference of second orbit
.
= 2 r2
2
2
(0.529 10
m
10
22 )
13.30 10 10 m
1840 me ]
h
Number of revolutions = (d)
1
RH Z 2 1
2 3eVmLi
He
He
h 2 3eV
1
13.6 32
E
2
[m p
3) from first excited
.
in z-direction = m 19.
1 3 3
9m p
3 h ; 4
o)
Hence, IP
20.
Li
h(
state, n1
x 1840
Hence, v (d)
ZH n
2eVm p
For the ionization of Li 2 ( Z
n
(b) Speed of electron in nth orbit, v
K .E .
(c)
1
h 2 (n = 2 for first excited state) (c) In the ground state electronic transition will occur only when an electron from K-shell i.e., first shell is captured by the nucleus (K-electron capture). Electron so captured converts proton into neutron resulting in the decrease of atomic number. (a) There is only one type of transition from n = 2 to n = 1 and hence emitted radiation will constitute only one frequency. (b) K.E. of an electron in a Bohr orbit is equal to the magnitude of the total energy but of opposite sign. So it varies inversly to the square of principal quantum number.
vH v
(d)
m
work function h
3
p
15.
300 10
19
(a) Angular momentum, mvr
He
14.
3 108
3 108
4.0 1.6 10
Hence,
13.
34
(a) Threshold frequency o =
6.
34
6.63 10
or 5.
6.63 10
Li
Hence,
9m p
RH
1
1
n12
n22
RH
22
1
1
2
2
1
2
1.095 10 13.3 10
2 10
8.23 10 6
; 1
1
2
42
2
1 H
for n2
2 to n1
1
STRUCTURE OF ATOM 21.
43
hcNA
(b) Energy of 1 mol of quanta = hvNA
30.
(a) For the transition n = 1 to n = 2 the energy change, E , is positive, i.e., energy is absorbed. For the transition n = 10 to n = 1, E is larger but of negative sign, i.e. energy is released.
31.
(b) Angular momentum m r
6.63 10 34 3 108 6.023 1023
472.2 kJ 253.7 10 9 Energy converted into kinetic energy = 472.2 – 430.53 kJ % of radiant energy converted into kinetic energy ( 472.2 430.53) 100 = 472.2
22.
(c)
h
p
2eVm p
;
2 2eVm
He
2
h 2 2eV p
Hence
He
23.
(c)
p h 4
d
h l (l 1) ;l 2
28.
(b)
max
RH
min
h 2(2 1) 2
n12
(n2
)
(d)
Li
,Z
Hence,
1 max
or
max
2; n2
3
RH Z 2
1
1
n12
n22
1 2 1 3 x 22
1
4x 5
2
3
m
4.
h 2 r4
5 4x
42
16a0
4h 2 16a0
h 8 a0
a0
8 a0
46.
(b) Number of radial nodes = n l 1
47.
(c)
49. 9 5 x 36
a0
h m
48.
3, n1
n2 , r4
rn
h( h(
(KE)1 (KE) 2 v1
RH
2
4s
41.
For largest wavelength in the Balmer series of 2
3d
40.
For H-atom, Z = 1; For Lyman series, n1 = 1
1 Hence, x
v
[Ar] 3d6 4s2 or [Ar] (a), (b), (c) - 3 unpaired electrons (d) - 4 unpaired electrons (c) 7s orbital, (n + l) is minimum with n minimum.
1 for
p-orbital. (b) Number of angular nodes l = 2 (l = 2 for d-orbital) Orbital angular momentum
Z2
nm
(a) For g-subshell, l 4 . Hence, n = 5 (minimum) Number of orbitals in 5th shell = n2 = 52 = 25 (a) Use (n + l) rule. (d) (a) [Ar] 3d3 4s2 (b) [Ar] 3d10 4s2 4p3 (c) [Ne] 3s23p3
2evm
(d)
(d) Orbital angular momentum =
1
h m
6
24.
h l (l 1) 2
1.23
h
h
33. 35. 39.
1 h m
(For first
(c)
h
2
h 2
32.
2 2
or (md )2
29.
h h 2 r circumference Hence, = circumference or, m
2
h 4
x p
4m p
h 2
orbit)
8.82%
h
He 2
n
50.
(c)
RZ 2
1
0)
2
0)
3
4.0 1015 Hz; 1
1
n12
n22
2
2.0 1015 Hz
For the last line of Lyman series of H-spectrum, Z = 1, n1 = 1 and n2 = For H line in Balmer series of He+, Z = 2, n1 = 2 and n2 = 3 (b) Z for He+ = 2, first excited state for He+, n = 2. Z for Li2+ = 3; Second excited state, n = 3 En = –13.6 Z2/n2 (c)
Cr 3 [Ar]3d 3 , U.P. 3; Fe 2 [ Ar ]3d 6 , Unpaired electrons = 4; Ni2+ – [Ar]3d 8, Unpaired electrons = 2 Cu2+ – [Ar] 3d 9, Unpaired electrons = 1
IIT-JEE CHEMISTRY Challenger
44
51.
(b)
n(n 2) electrons)
3 (n = number of unpaired
15, n
Therefore, x = 4 52.
(d)
Fe
2
[Ar]3d
M
4
[Ar]3d
69.
=
6 ; number of d-electrons = 6
6
2
Cl(17) 1s 2 s 2 p 3s 3 p 53.
(b)
70.
5 ; p-electrons = 11
71.
Fe(III) [Ar]3d 5 ; unpaired electron = 5; magnetic 72.
5(5 2) B.M.
moment =
CO(II) [Ar]3d 7 ; unpaired electrons = 3; magnetic moment = 55. 56. 57.
3 (3 2) B.M.
74.
59. 60.
(d)
(b)
En
l
hcRZ
En(3)
1 2
78.
9 1 10 ; Maximum multiplicity = 2S + 1 = 2 2 minimum multiplicity = 2
13.6
63.
(d) Magnetic moment
65.
=2 6 24 B.M. n(n 2) B.M. Hence, n = 4 (unpaired electrons) Co3+ – [Ar] 3d6, 4 unpaired electrons. (c) Number of nodal surfaces = n l 1 3 0 1
68.
(c)
79.
n2 = 4 ............. (i)
n2
= RH
32
Z2
; Z for Li 2 = 3
2
13.6eV
32
1 × Potential energy 2
(d) K.E. of electron = –
Ze 2 r
1 2
Z2
= 13.6
Li 2
2
13.6
Z2
13.6
n2
32 12
n12
82.
n22 1
1
1
2
1
3
8 RH
122.4eV
Total energy = – 122.4 eV = P.E. + K.E. = P.E.
n1 = 2 ........... (ii)
1
22
= 54.4 eV 12 n2 Hence, the total energy required for the removal of both the electrons = 24.6 + 54.4 = 79.0 eV (c) Total energy of electron in the ground state (n = 1) for
P. E. = – 244.8 eV
1
1 3e 2 2 r
(b) Electron in lower energy state move faster. Hence, K.E. increases during transition from higher to lower state. (d) Energy required for the removal of second electron
n1 = 1, n2 = 3 = RH Z 2
32
from He-atom = 13.6
1 1 2 2
19
30.6 eV
22
=– 77.
1 2
Minimum total spin =
n1
32
13.6
4 ; number of degenerate orbitals = 2l 1 9 ;
maximum total spins = 9
1.6 10
n n for the excited states = 2, 3, 4 etc.
En(2)
75.
31
(c) Energy of electron = – 13.6
2
hcR ( Z 2) 2 4 n n = 4, number of orbitals in 4th shell = 4 2 = 16 (d) ml l is not permissible.
58.
2meV
2 2 25 50 5 2 25 , No. of electrons 2 n (d) Number of spherical or radial nodes = n – l – 1 =5–4–1=0 Number of non-spherical or angular nodes = l = 4 (c) For 3p-orbital, n = 3, l = 1 No. of spherical nodes = n – l – 1 = 3 – 1 – 1 = 1 No. of non-spherical nodes = l = 1
2.19 106
v
34
h
V = 150 volt V (d) For ‘g’ subshell, l = 4. The minimum value of principal quantum number n = 5. No. of orbitals in 5th shell
Z ms 1 or vn 2.19 106 Z ms 1 n (c) Two electrons in the same orbital can not have the same spins. (d) Electrons in degenerate orbitals can not pair up until they are singly occupied. (c)
6.63 10
2 9.1 10
Ne(10) 1s 2 2s 2 2 p 6 ; p-electrons = 6 2
h mv
3
Mg(12) 1s 2 2 s 2 2 p6 3s 2 ; number of s-electrons = 6
2
m=
10
= 1Å = 10
(b)
(c)
E
13.6 Z 2
13.6 3 2
1 2
2
1
1
n12
n22 1 2
eV
= 30.6 eV
P.E. 2
STRUCTURE OF ATOM 83.
(a)
=
1
45
= RH
1
1
n12
n22
For the largest wave length in Balmer series of Li2+ the transition occurs from n2 = 3 to n2 = 2 and in such a case we have
1
= 1.097 10 7
1
1
2
2
1
max
91.1 10 9 m
85.
(d) Energy for photon E 34
6.63 10
3 108
J
66.3 10
19
300 10 10 K.E. acquired by the electron
= 66.3 10
19
1 1.6 10
1 2 1 9.1 10 mv = 2 2 v
86.
31
19
J
= 64.7 10
=
19
J
6.63 10
34
=
3 108
4 1.6 10
310 10 m 310nm 2
2
4
87.
(b)
90.
= 2 1.1 2 1.1 – 4 × 7 = – 23.6 MeV (b) A Positron has characteristics the same as an electron except for the sign of its charge.
1H
Hence, En
2 He
13.6
Z2
n(n 1) 2 (d) For the shortest wave length of spectral line in He+ the transition should be from n n1 N
1 min
or
1 x
or RH
RH .
RH .
Z2
4 [For He+; Z = 2, 1
1 ....... (i) 4x
or
min = x]
1 2
(2) 1
( )2 1
(2) 2
(3)2
RH
1 4x
1 0 4 9 4 36
1 4
36 5
9 5
9:5
1 , we get K.E. l = and K.E.1 = E (initial) 2 = 0.9
(d) Using the relation,
[10% decrease from
1,
2=
–
10 100
= (1 – 0.1) = 0.9 ] We are required to find K.E.2 = ?
K.E.2
n12
1 9
16 x 5 (b) In the Balmer series of hydrogen atom the highest wave-length is that of first line and the least wave length is that of limiting line, thus
1
94.
(3)2
max
1
12
(2)
1 2
5 16 x
max
Energy change
= 13.6 2 = 3.4 eV n2 2 (for 1st excited state, n = 2) (a) We know that number of spectral lines in the emission spectrum of hydrogen atom, in which the electron jumps from nth energy level to ground state (n = 1) is given by the relation
Thus
1
or
19
1
9 5 4 x 36
max
or
hc
n22
9 9 4 4 x 36
1
or
93.
92.
1 max
v2
1
n12
1 1 9 4x 4
max
or
9
91.
1
or
1
RH .(3) 2
max
3.8 106 ms 1
(c) Energy of photon, E = h
1H
1
or
hc
h
RH .Z 2
2 1 2 2
K.E.1
2
100
2
81 100 E 81
2
(.9 )2 E
K.E.1
[K.E1 = E]
[more than K.E.1 i.e., increase]
IIT-JEE CHEMISTRY Challenger
46
100 E E 81
Now increase in K.E. = K.E.2 – K.E.1
or
% increase of K.E.
or [
1.23 10 10
m
100
23.4%
m, we get
V
9
1.23 10
1.23 × 10–10 m
or
9
123 × 10–12 m = 123 pm
h 2meV 100.
6.63 10 34
( 34
6.63 10 25
15.4 10
t1 t2 97.
98.
100
100
(2) (3)
3
or
hc
hc
hc
1
2
3
1 2
3
1
or
n3, we get
1
1
1
2
3
2
1. 2
1 3
(d) From the given data, when the system moves from 2E level to E level, we have 2E–E or
E
or
E 3
hc
or
hc .3
hc
or
1 3
1
1
hc
E
1
1
or
1
3
or
1
3
1 2 mv 2
Thus E
1 2 mv or v 2
2E m
Also
h mv
h
h
2E m m
2mE
103.
(b) It depends upon quantum numbers l, m.
104.
(a) We know
h and K.E. mv
1 2 mv 2
From the above relation, we get K.E.
hv 2
K.E.
hc
mv hc
is the wave length of
(c) We know K.E.
m]
1
1 2
1
102.
9
8 27 or 8 : 27
3
[
1
101.
V = 100)
(a) In case of a ‘5d’ orbital, we have, n = 5, l = 2 Thus, number of nodal planes = l = 2 number of nodes = n – l – 1 = 5 – 2 – 1 = 2 number of peaks = n – l = 5 – 2 = 3 (b) From the given data, we have (EC – EB) + (EB – EA) = (EC – EA) or
99.
1.23 10
(c) Using the relation, time period (t)
n13 n23
hc
(d) In a hydrogen like atom, the difference in energy between successive levels decreases with increasing values of n. Since the transition from n = 4 to n = 3 results in ultraviolet radiations, the infra-red radiations (which involve smaller amounts of energy as compared to ultraviolet rays) will be obtained in a transition that involves initial values of n greater than 4. The only such option is transition form n = 5 to n = 4. (a) It depends upon quantum numbers n, s.
(E = eV)
2 9.1 10 31 1.6 10 19 100
96.
4 E E 3
photon emitted]
1.23 10 9
(a) Using the relation,
4E level to E level, 3
we have
100 E E 81 100 E
19 100 81 95.
When the system moves from
1 2 mv 2 2K.E. ; v
hv or K.E. 2K.E.
Since K.E. for two particles is same, so i.e., option (a)
v
hv 2
STRUCTURE OF ATOM 105.
47
(b) The energy associated with radiation of wave length 975 Å is given by
12400 eV 975
E
or
1. 2.
13.6
E
12400 eV] x
6. 7.
8.
n12
n22
eV
[
n1 = 1]
1
h
2me 5000
100
me
13.
2h . Hence r 2
m
34
27
10
200
h 2
9
m 10Å
14.
s(s 1) is determined
by the value of spin quantum number. (c) In one electron species, the energy of an orbital depends only on the value of n. Hence, 3s, 3p, and 3d orbitals are of equal energy (degenerate orbitals). (b) In (b), 2s orbital of lower energy is not completely filled before the filling of 2p-orbital of higher energy.
h
1.23 10 9
2 eVm
V
13.6 12.72 13.6
or
n22
16
or
n2
4
11.
(c)
12.
(d) Number of angular nodes = ‘l’ (or non spherical nodes) = 4 [For 5g orbitals l = 4] Number of radial nodes = (n – l – 1) (or spherical nodes) = 5 – 4 – 1 = 0 [For 5g, n = 5]
m
0.88 13.6
1 16
n3 will give the longest wave
hcR.Z 2
(d) We know, En
n2
hcR.4 n 4
2
hc.R 4
1 4 n or n2 = 16 or n = 4 = Number of sub-shells. (d) Given : subsidiary quantum number (l) = 4 number of degenerate orbitals = 2l + 1 = 2 × 4 + 1 =8+1=9
or
(d) Spin quantum number(s) had to be introduced to account for the splitting of spectral lines under the electric field (Stark effect). (d) Spin angular momentum =
n22
or
h mv
(d) de Broglie wavelength 6.6 10
E = 12.72 eV]
12.72 13.6
n22
1
[
n22
For He+, Z = 2 hcR.4 En n2
(c) The angular momentum of electron in second orbit is
h n 2
1
1
The transition n4 length.
h
3.3 10
or
or
1 1 eV 1 n22
2meV
h mv
5.
1
h
given by mvr 4.
1
(d) Momentum can not be calculated as the velocity of hydrogen atom is not given. (b) Wavelength of electron accelerated by potential of V volt =
3.
13.6
12.72 13.6 1
12.72 eV
[For a radiation of wavelength x Å, we know E
E
or
2
For this maximum total spins (s)
9
1 2
For this minimum total spins (s)
1
1 2
Maximum total multiplicity = 2s + 1 = 2 ×
1 +1=2 2 (c) Fe(Z = 26), Fe2+; 3d6, number of d-electrons in Fe2+ = 6 In Fe, 3d64s2, so number of d-electron = 6 In Ne(Z = 10); 1s2 2s2 2p6; number of p-electrons = 6 In Cl (Z = 17); 1s22s2p63s2p5, so number of p-electrons = 6 + 5 = 11 In Mg(Z = 12), 1s22s22p63s2, so number of s-electrons =2+2+2=6 Thus number of p-electrons in Cl is not equal to number of d-electrons in Fe2+. Minimum total multiplicity = 2 ×
15.
9 + 1 = 10 2
IIT-JEE CHEMISTRY Challenger
48
1.
2.
3.
4.
5.
6.
1. 4.
(b) K.E. = h ( 0 ) ; K.E. increases linearly with , the frequency. The rate of emission of photoelectrons i.e., the photoelectric current is proportional to the rate of impinging photons i.e. the intensity of light. (d) The threshold frequency is the minimum frequency required for the ejection of electron from the metal surface. (c) Plus and minus signs of spin quantum numbers imply that spin angular momentum of the electron, a vector quantity, acts in the same or opposite directions of orbital angular momentum. (d) In hydrogen and hydrogen like species (one electron species), energy of electron is determined only by principal quantum number, not by n and l. Hence, 3s and 3p are degenerate orbitals i.e., have equal energies. (c) Electromagnetic waves consist of oscillating electric and magnetic fields and radiate into the space. de Broglie or matter waves are always associated with the matter particles (electrons) and do not radiate in space. (b) The difference between the energies of adjacent energy levels decreases as n value increases. Thus, for H-atom
(b,c) (a,c,d)
(b) l n is not permissible; (c) m permissible. Angular momentum, m r
n.
l is not
E2
(a,b,c,d) En
K.E. n
6.
7.
(a,d)
(b)
Z 13.6 2 ; for He , Z n En
13.6
z2 n2
;
2.19 106 Z / n ms
1
8.
(d)
9.
(c)
10.
(a)
11.
(b)
12.
(b)
13.
(b)
14.
(b)
8.
(c)
2 px and 2 pz orbitals are degenerate i.e., they are of equal energies and hence no transition of electron between them is possible. Raising the electron to infinity implies its removal from the atom. He2+ has no extranuclear electron. Except Li, elements of first group have low ionization energy. At the temperature of the flame, when the excited electrons return to the ground state, energy in the visible region is emitted. Both assertion and reason are correct. Reason is not the correct explanation of assertion. Both assertion and reason are correct. Reason is not the correct explanation of assertion. Both assertion and reason are correct. Reason is not the correct explanation of assertion. Both assertion and reason are correct. Reason is not the correct explanation of assertion.
Zn2+ and Cu+ have no unpaired electrons and hence diamagnetic. O 22
has also no unpaired
electron (follow M.O. theory) and hence diamagnetic. He 2 has one unpaired electron (paramagnetic). 10.
(b,c,d)
Li2+, D and T are one electron species. He2+ has no electron.
13.
(a,c)
There is some mass defect due to binding energy. Hence (c)
16.
(b)
A – 1s2 2s22p 63s23p5 – Cl (17), 11 p electrons B – 1s22s22p63s23p64s23d1, (Sc - 21), l = 2 and ml = 2 for 3d-electron.
(a) Both Fe3+ and Mn2+ have same number (5) unpaired electrons. (b) Cu2Cl2 : Cu (I) has no unpaired d-electron, hence colourless. CuCl2 : Cu(II) has one unpaired d-electron, hence coloured. Atomic number z of M = 18 (atomic number of Ar) + 1 = 19 Number of neutrons = 19 + 1 = 20 ; mass number =p+n
E 3 .....................
E4
(d) The orbital has dumb-bell shape and belongs to psubshell.
h , n is an integer.. 2
2
E3 E 2
7.
2
5.
E1
C – 1s22s22p63s23p64s23d2 (Ti - 22), n = 3 and l = 2 for 3d 2 electrons. 18.
(b, c)
Statements (a) and (d) are incorrect. (a) The excited electron has a tendency to return back to lower energy state. (d) Angular momentum
nh 2
STRUCTURE OF ATOM
1.
49
A-q, B-r, C-s, D-p 2
2
kZe , rn
Vn
rn
(A)
(B)
kZe , 2rn
Kn
But E1
2
En
Vn
Kn
kZe , 2rn
(C)
(C) rn
kZe2 1 En 2
kZe2 , rn 2rn En 1 , Hence x kn 2 1 , rn Z
5.
Z
1 , 2 rn kn
n1
h 0(0 1) =0 2
2eV = me
= 10
e ms me
1
(D) Wavelength
h ; me v
n2
1
13.6Z 2 n
2
n1
1
1
1
n12
n22
max )
in Lyman series
1
kZ 2
1
2
k
2
1
1 x
1 2 1 3 2 x 1
1 2
=
9 x
2, n2
Hence, 1
1 1 4x 2 2 9 2 (D) For longest wavelength in Balmer series, n1 = 2, n2 = 3
10 eme
h 10 eme
1
13.6 22 =
kZ 2
1 2 1 1 27 4x 3 2 2 x 4 x 27 1 2 (C) For shortest wavelength in Balmer series,
6.
(A) For He , Z = 2; for ground state of He , n = 1 Hence, E1
1
2
54.4 eV
1
1 x
2
3
1 2 1 1 4x 3 2 2 x 5 2 3 A-p, q, r, s; B-q, r, s; C-s; D-r, s For Lyman series, n1 = 1 so n2 = 2, 3, 5, 6 For Balmer series, n1 = 2 so n2 = 3, 5, 6 For Pfund series, n1 = 5 so n2 = 6 For Bracket series, n1 = 4 so n2 = 5, 6 Hence,
eV atom
= 13.6 eV atom–1
1
A-r; B-s; C-p; D-q
En
2
n2 2
x 9
(B) For longest wavelength in Lyman series, n1
2eVme 2e 50 M e
1
1
n12
2,
and Z = 1 (for H-atom) 1 x
Hence
2 50 e me
2evme
Momentum mev =
2
2
1
n1
(A) For shortest wavelength in Lyman series of Li 2 ,Z=3 n 1, n
(B) K.E. = eV =50e J
4.
1 , n2
Hence,
1 2 K.E. of electron = me v = P.E. of electron = eV 2 (A) Velocity of electron =
1
For shortest wavelength (
A-s; B-r; C-q; D-p
=
54.4 eV atom–1
A-r, B-s, C-q, D-p Wave number v
Z Hence, y = 1
h l (l 1) = 2
=
(C) me2v 2
E 13.6 Z 2
2 = 13.6 2
1
54.4
54.4 eV atom–1
Etotal
Hence, I.E. =
(D) 1s is the lowest orbital, for which l 0 Angular momentum of electron in lowest orbital
2.
E1
( 108.8) 2
(D) For ionization of 1st excited state of H e n2
2
kZe2 2rn
En rn
P.E. 2
(B) K.E.
Vn Kn
P.E. 2
P.E. = -108.8 eV
kn 2 Z kZe2 rn
P.E. 2
P.E. K.E. = P.E.
1,
IIT-JEE CHEMISTRY Challenger
50
1.
Ans : 3.31 Energy required to excite an electron from n1 energy level to n2 energy level in H-atom like species is given by
E = KZ 2
1 n12
–
1 n22
3.
÷
K.E. of photoelectron
Hence
He+,
In case of
( E )3
1 n12
= –13.6 2 2
–
1 n22
1
–
2
8 6.02 10 9 2
– 54.4
– 2 × 1015 =
1 2
( E )2
He+
for 25% of
1
1
–54.4
1
–
2
1
2
2÷
÷ eV atom–1
4.
1
2
–
6.02 10 23 eV 4
91 6.02 10 23 1.6 10 –19 J 144 = 3.31 × 103 kJ = –54.4
v= 1
= RZ
2
2
R 1
H
1
n12
–
1
– 34
÷ 1500 10 –10 ÷ – 2 × 1015)
= 72.53 1015
n12
–
1
1
– RZ 2
2
–
n12
1
Na +
n22
1 2
–
( n1 + 1)2 109677.7 22
n1 = 3
( n1 + 1)2
Wavelength of first line in the spectral series 1
=
1
= 109677.7 2 2
3
2
–
1
= 4689 10 –8 cm
42
= 4689 Å 5.
Ans : 6.57 (K.E.)1 = 1.8 eV = h (K.E.)2 = 4.0 eV = h
+ IE =
1
hc
+ IE
......(i)
1 2
+ IE =
hc
+ IE
......(ii)
2
1
–
2
÷ (4 – 1.8) eV = h × 3.0 × 108
1
– 1 2
h
1
363 ÷= 4 R 2 2
363R = 4 = 653.4 5R 36
÷
( n1 + 1)2 ÷
From (i) and (ii), (K.E.)2 – (K.E.)1 = hc
1
1 H
1
5 – 2÷= R 2 36 2 3
R 112
Na +
Hence
1
÷
3 108
Ans : 653
1
c
RZ 2
2.7451 104 =
ions
4.806 10 –17
1
= RZ 2
=
3 – 54.4 6.02 1023 eV 16
2.
–
eV
4 3 + ÷ 6.02 10 23 eV 9 16
–
6.626 10 = 7.45 × 1016 Hz Ans : 4689 Extreme lines mean first and the last (convergence limit). Let n1 be the energy level of He+ ion.
23
Total energy change = –54.4
=h
= Threshold frequency)
= 6.626 × 10–34 (
4 6.02 1023 eV 9
– 54.4
0)
4.806 × 10–17 = 6.626 × 10–34
for half mole of He+
1
0
÷÷
1 3 8 eVatom–1 –54.4 9
( E )3
(
1 1 – 2 ÷ = K (Z = 1) 12
Z=2
= – KZ 2
1
h( –
0
For ionization of H-atom, n1 = 1 and n2 =
E = 13.6 eV = K 12
Ans : 7.45 K.E. of photoelectron = e × stopping potential = 1.602 × 10–19 × 300 = 4.806 × 10–17 J
h=
2
1 1
÷÷
÷÷
3.0 108 (800 – 700 ) 10 –10 700 10 –10 800 10 –10
(2.2 1.6 10 –19 J) 700 10 –10 800 10 –10 m 2
3.0 108 (ms 1 ) 100 10 –10 (m) = 6.57 × 10–34 Js
