2a.atomic Structure( 41-64 )

JEE ADVANCED - VOL - I

ATOMIC STRUCTURE

ATOMIC STRUCTURE Illustration: The wavelengths of the characterstic K X-rays of iron and potassium are 1.932 x 10-8 and 3.737 x 10-8cm, respectively. What is the atomic number of an element for which the characterstic K wavelength is 2.289 x 10-8cm? Solution:From Moseley’s law

SYNOPSIS Mosley’s Experiment:By using different X-ray tubes provided with anti-cathodes of different materials, Moseley was able to take the spectrum of Xrays in each case. It was observed that the wavelengths of X-ray were characterstic of each element. The characteristic X-ray spectrum consists of discrete spectral lines which can be ground into K-series, L-series, M-series,etc.

 Z (or )   aZ 2

 

K

M

L

 108

Z

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Moseley analysed the result as ‘ the fast moving cathode rays were able to remove electrons from the inner orbits of an atom of anti-cathode due to collision.

K and K  lines, b = 1. Hence

  a ( Z  b) where ‘a’ is the proportionality constant. This equation is very useful for the calculation of Z if the frequency of K and K  lines are known. Narayana Junior Colleges

1 (26)2

For K, 2

1 (19)2

For X, 3

1 ( Z )2

1.931 108 Z   (26)2 8 2.289  10 2

The results obtained led to the suggestion that  must be directly proportional to the atomic number of an element (Z)

 ( Z  b) Where ‘b’ is the screening constant, for

For Fe, 1

1 ( Z )2  Now 3 (26)2

Moseley showed that the square root of the frequency of a spectral line is strictly related to the nuclear charge (Z) if the excitation potential is fixed.

 Z To give accurate results, Moseley modified this equation as

c 1  2 2 aZ Z

*

Z  23.88  24 Electromagnetic Spectrum Electromagnetic radiation is not a single wavelength radiation, but a mixture of various wavelength or frequencies. All the frequencies have same speed. If all the components of Electromagnetic Radiation (EMR) are arranged in order of decreasing or increasing wavelengths or frequencies, the pattern obtained is known as Electromagnetic Spectrum. The following table shows all the components of light. 41


Wavelength Frequency Source (nm) (Hz)

1.

31014  3 107 1  1 0 5  1  1 0 9 Alternating

Radio wave

current of high frequency Micro 3 107  6106 1  10 9  5  1011 Klystron wave tube Infrared 6  10 6  7600 51011 3.951016 (IR) Incandescent objects Visible 7 6 0 0  3 8 0 0 3.951016 7.91014 Electric bulbs, sun rays Ultra violet 3800-150 7.9  1014  2 1016 Sun rays, arc lamps with (UV) mercury vapours X-rays 150-0.1 2 1016  3 1019 Cathode rays striking metal plate   rays 0.1-0.01 3  1019  3  10 20 Secondary effect of radioactive decay Cosmic rays 0.01-zero 3  1 0 2 0 -infinity Outer space

2. 3. 4. 5.

6. 7.

8.

Continuous Spectrum: When sunlight (white light) is passed through a prism, it is dispersed or resolved into a continuous spectra of colours. It extends from RED (7600 Å) at one end to the VIOLET (3800Å) at other end. In this region, all the intermediate frequencies between red and violet are present. The type of spectrum is known as Continuous Spectrum., Hence continuous spectra is one which contains radiation of all the frequencies. Discontinuous Spectrum: Light emitted from atoms heated in a flame or excited electrically in gas discharge tube, does not contain a continuous spectrum of wavelengths (or frequencies). It contains only certain well-defined wavelength (or frequencies). The spectrum pattern appears as a series of bright lines (separated by gaps of darkness) and hence called as Line-Spectrum. One notable feature observed is, that each element emits a characteristic spectrum, suggesting that there is discrete relation between the spectrum characteristics and the internal atomic structure of an atom. PHOTOELECTRIC EFFECT It was observed by Hertz and Lenard around 1880 that when a clean metallic surface is

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42


S. Name No.

ATOMIC STRUCTURE

irradiated by monochromatic light of proper frequency, electrons are emitted from it. This phenomenon of ejection of the electrons from metal surface was called as Photoelectric Effect. It was observed that if the frequency of incident radiation is below a certain minimum value (threshold frequency), no emission takes place however high the intensity of light may be. Another important feature observed was that the kinetic energy of the electrons emitted is independent of the Intensity of the light. The kinetic energy of the electrons increases linearly with the frequency of incident light radiation. This was highly contrary to the laws of Physics at that time i.e. the energy of the electrons should have been proportional to the intensity of the light, not on the frequency. These features could not be properly explained on the basis of Maxwell’s concept of light i.e. light as electromagnetic wave. In 1905, Einstein applied Planck’s quantum theory of light to account for the extraordinary features of the photoelectric effect. He introduced a new concept that light shows dual nature. In phenomenon like reflection, refraction and diffraction it shows wave nature and in phenomenon like photoelectric effects, it shows particle nature. According to the particle nature, the energy of the light is carried in discrete units whose magnitude is proportional to the frequency of the light wave. These units were called as photons (or quanta). According to Einstein, when a quantum of light (photon) strikes a metal surface, it imparts its energy to the electrons in the metal atom. In order for an electron to escape from the surface of the metal, it must overcome the attractive force of the nucleus in the metal atom. So a part of the photon’s energy is absorbed by the metal surface to release the electron, this is known as work function of the surface and is denoted by  . The remaining part of the energy of the photon goes into the kinetic energy of the electron emitted. If E is the energy of the photon, KE is the kinetic energy of the electron and  be the work function of the metal then we have; Narayana Junior Colleges


ATOMIC STRUCTURE

  h 0 and Ei  h

KE  Ei – 

Note: The electromagnetic radiation (or wave) now emerges as an entity which shows dual nature i.e., sometimes as Wave and sometimes as Particle (quantum aspect). Illustration 2 : In a photoelectric experiment, the collector plate is at 2.0 V with respect to emitter plate made of copper (work function 4.5 eV). The emitter is illuminated by a source of mono-chromatic light of wavelength 200 nm. Find the minimum and maximum kinetic energy of photoelectrons reaching the collector. Solution: Since plate potentials 2 V, minimum K.E. will be 2 eV. For max. K.E. use the following relation: Absorbed energy = Threshold energy + K.E.



KE  h – h0  h( – 0 )



Also, if m be the mass and v be the velocity of the electron ejected then KE 

KE 

1 2

mv 2  h( – 0 ) .

c c  1 2 mv  h  v  v0   h    2   0 

1 1       hc     hc  0    0   .0 

KE

Intensity of light

KE v 0  Threshold frequency


hc  4.5  1.6  10 –19  K.E.  6.626  10 –34  3  108  4.5  1.6  10 –19  K.E. –9 200  10

*

2 r 2 r nn  h \ mv 2 mvr 2 nh n n  h h 2 

Frequency

* Photo electric current

* Intensity

Photo electric current

*

Frequency


K.E. = 2.739 10 –19 J  1.7 eV Max K.E. = 2eV + 1.7 eV = 3.7 eV. No. of waves per revolution made by an electron in ‘n’th orbit is

QUANTUM NUMBERS To understand the concept of Quantum Numbers, we must know the meaning of some terms clearly so as to avoid any confusion. Energy Level: The non energy-radiating circular paths around the nucleus are called as Energy Levels or Shells. These are specified by numbers having values 1, 2, 3, 4, ... or K, L, M, N, ... in order of increasing energies. The energy of a particular energy level is fixed. Sub-Energy Level: The phenomenon of splitting of spectral lines in electric and magnetic fields reveals that there must be extra energy levels within a definite energy level. These were called as Sub-Energy Levels or Sub-Shells. There are four types of sub-shells namely; s, p, d, f. 43


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44

When n = 1,  = 0, i.e., its energy level contains one sub-shell which is called as a s-sub-shell. So for  = 0, the corresponding sub-shell is a ssub-shell. Similarly when  = 1, 2, 3, the subshells are called p, d, f sub-shells respectively. As you know for n = 1,  = 0, there is only one sub-shell. It is represented by 1s. Now for n = 2, l can take two values (the total number of values taken by  is equal to the value of n in a particular energy level). The possible values of  are 0, 1. The two sub-shell representing the IInd energy level are 2s, 2p. In the same manner, for n = 3, three sub-shells are designated as 3s, 3p, 3d corresponding to  = 0, 1, 2, and for n = 4, four sub-shells are designated as 4s, 4p, 4d, 4f corresponding to  = 0, 1, 2, 3. The orbital Angular momentum of electron = (   1) Narayana Junior Colleges

First energy level (K or 1) has one sub-shell designated as 1s, the second energy level (L or 2) has two sub-shell as 2s & 2p, the third energy level (M or 3) has three sub shell as 3s, 3p and 3d, and the fourth energy level (N or 4) has four sub-shells as 4s, 4p, 4d and 4f. The energy of sub-shell increases roughly in the order: s < p < d < f. Orbital: Each sub-energy level (sub-shell) is composed of one or more orbitals. The orbitals belonging to a particular sub-shell have equal energies and are called as degenerate orbitals. s-sub-shell has one orbital, p has three orbitals, d have five orbitals and f has seven orbitals. To describe or to characterize the electrons around the nucleus in an atom, a set of four numbers is used, called as Quantum Numbers. These are specified such that the states available to the electrons should follow the laws of quantum mechanics or wave mechanics. Principal Quantum Number: (n): This quantum number represents the main energy levels (principal energy levels) designated as n = 1, 2, 3, ... or the corresponding shells are named as K, L, M, N, ... respectively. It gives an idea of position and energy of an electron. The energy level n = 1 corresponds to minimum energy and subsequently n = 2, 3, 4, ..., are arranged in order of increasing energy. Higher is the value of n, greater is its distance from the nucleus, greater is its size and also greater is its energy. It also gives the total electrons that may be accommodated in each shell, the capacity of each shell is given by the formula 2n2 , where n : principal quantum number. Azimuthal Quantum Number (  ): This number determines the energy associated with the angular momentum of the electron about the nucleus. It is also called as the angular momentum quantum number. It accounts for the appearance of groups of closely packed spectral lines in electric field. It can assume all integral values from 0 to n–1. The possible values of  are : 0, 1, 2, 3, ..., n–1. Each value of  describes a particular sub-shell in the main energy level and determines the shape of the electron cloud.

ATOMIC STRUCTURE

h . 2

Note that its value does not depend upon value of n. Magnetic Quantum Number (m): An electron with angular momentum can be thought as an electric current circulating in a loop. A magnetic field due to this current is observed. This induced magnetism is determined by the magnetic quantum number. Under the influence of magnetic field, the electrons in a given sub-energy level prefer to orient themselves in certain specific regions in space around the nucleus. The number of possible orientations for a sub-energy level is determined by possible values of m corresponds to the number of orbitals in a given sub-energy level). m can have any integral values between –  to +  including 0, i.e., m = –  , 0, +  . We can say that a total of (2  + 1) values of m are there for a given value of  . In s sub-shell there is only one orbital [  = 0, m = (2  +1) = 1]. In p sub-shell there are three orbit als corresponding to three values of m : –1, 0 +1. [ = 1



ATOMIC STRUCTURE

 m = (2  +1) = 3]. These three orbitals are represented as p x , p y , p z along X, Y, Z axes

n = 6,  = 0, 1, 2, 3, 4, 5, m = 0, = 1, m = –1, 0, +1,  = 2, m = –2, –1, 0, +1,  +2,  = 3, m = –3, –2, –1, 0, +1, +2, +3,  = 4, m = –4, to +4 including 0,  = 5, m = -5 to +5 including zero. b)(i) 2p, (ii)3s

perpendicular to each other representing three orientations. In d sub-shell, there are five orbitals corresponding to –2, –1, 0, +1, +2, [  = 2  m = ( 2  2  1)  5] . These five orbitals are represented as dxy ,dyz ,dzx ,dx2 –y2 ,dz2 .

2p y2

2

1

1

+1/2, -1/2

1 0 +1/2, -1/2 2p z 2 2 Illustration 4 : a) An electron is in 6f-orbital. What possible values of quantum numbers n,  , m and s can it have? b) What designation is given to an orbital having (i) n = 2,  =1, and (ii) n = 3, = 0? Solution: a) For an electron in 6f-orbital, quantum number are: Narayana Junior Colleges


In f sub-shell t here are seven orbitals corresponding to –3, –2, –1, 0, +1, +2, +3 [  = 3  m = (2  3  1)  7] . * Spin quantum Number (s): When an electron rotates around a nucleus it also spins about its axis. If the spin is clockwise, its spin quantum number is +1/2 and is represented as  . If the spin is anti-clockwise, its value is –1/2 and is represented as  . If the value of s is +1/2, then by convention, we take that electron as the first electron in that orbital and if the value of s is –1/ 2, it is taken as second electron. Spin quantum number do not depend on other quantum numbers. The quantum numbers +1/2 and -1/2 for the electron spin represent two quantum mechanical spin states which have no classical analogue. Illustration 3 : Write down the values of quantum numbers of all the electrons present in the outermost orbit of neon (At. No. 10) Solution: The electronic configuration of neon is1s2, 2s2,2p6. Values of quantum numbers are: n m s  0 0 +1/2, -1/2 2s2 2 1 +1/2, -1/2 2p x 2 2 1

* i)

SHAPES OF ATOMIC ORBITALS s-orbital: An electron is considered to be immersed out in the form of a cloud. The shape of the cloud is the shape of the orbital. The cloud is not uniform but denser in the region where the probability of finding the electron is maximum. The orbital with the lowest energy is the 1s orbital. It is a sphere with its center of the nucleus of the atom. The s-orbital is said to spherically symmetrical about the nucleus, so that the electronic charge is not concentrated in any particular direction. 2s orbital is also spherically symmetrical about the nucleus, but it is larger than (i.e., away from) the 1s orbit. y

2s

1s x

nucleus

Z

ii)

radial node

p-orbitals: There are three p-orbitals: p x ,p y and p z . they are dumb-bell shaped, the

two lobes being separated by a nodal plane, i.e., a plane where there is no likely hood of finding the electron. The p-orbitals have a marked direction character, depending as whether p x ,p y and p z orbital are being considered. The

p-orbitals consist of two lobes with the atomic nucleus lying between them. The axis of each porbital is perpendicular to the other two. The p x ,p y and p z orbitals are equivalent except for

their directional property. They have same energy; orbitals having the same energy are said to be degenerated. 45


ATOMIC STRUCTURE

Y 'Y Z ' is n o d a l p la n e

* X 2 P x o r b ita l

Electronic configuration of an element is represented by the notation nl x . x : number of electrons present in an orbital, l : denotes the sub-shell n : principal quantum number. Radial distribution function : The function which determines the probability of finding an electron at a distance “r” from nucleus is called “Radial distribution function”.

Z z

z

z y

y

y

x

x

x

+

iii)

py

pz

d-orbitals: There are five d-orbitals. The shapes of four d-orbitals resemble four leaf cloves. The fifth d-orbital looks different. The shapes of these orbitals are given below. y

z

z

y

x d xy

dyz

y

d x 2 y 2 *

dzx

z

x

x

x

dz 2

Extra Stability of Half and fully Filled Orbitals: A particularly stable system is obtained when a set of equivalent orbitals (degenerate orbitals) is either fully filled or half filled, i.e., each containing one or a pair of electrons. This effect is more dominant in d and f sub-shells. This means three or six electrons in p-sub-shell, five or ten electrons in d-sub-shell, and seven or fourteen electrons in f-sub-shell forms a stable arrangement. Note this effect when filling of electrons takes place in d sub-shells (for atomic number Z = 24, 25, and 29, 30). 46


px

*

Quantum Mechanical Model of Atom Considering electron as a 3-D wave Schrodinger proposed a wave equation to de scribe the electron around the nucleus at a point

 2  2  2 8 2 m    2  E  U   0 x 2 y 2 z 2 h Where  (Psi) = (Wave function) amplitude of electron wave x,y,z = cartesian coordinates m = mass of electron h = planck’s constant E = permissible total energy U

=

  ze 2  potential energy  r   

Schrodinger wave equation can be trans formed into a wave equation in terms of the spherical polar coordinates  r, and   of the electron with respect to the nucleus. such that x  r sin  cos  y  r sin  sin 

z  r cos  x2  y 2  z 2  r 2 Narayana Junior Colleges


ATOMIC STRUCTURE

When wave equation in polar coordinates is solved it gives the possible energy states and corresponding wave function

Node 2

  r , ,   which is called atomic orbital

2s

  r , ,    Rn,l  r  l ,m   m       

r

angular part

Radial part

radial part dependent on ‘r’ and angular part dependent on  and   as a function of distance from the

nucleus(r): 1s



2s

 Node


r

Radial probability distribution : D-function (or) distribution function If we imagine infinitisimally small layers around the nucleus (a layer of thickness ‘dV’) the probability of finding the electron in each layer is given by radial distribution function. Consider atom of spherical volume... 4 V   r3 3

r

4 2 differentiating dv   3 r .dr 3

Amplitude   is not a characteristic prop erty of a wave to describe it completely. so

dv  4 r 2 dr

 2 is considered which is the intensity i.e. Probability of finding the electron in a 3D-

If multiplied by ‘ 2 ’ the probability at a point gives distribution of electron in the small layer Plot of ‘D’ as a function of distance:

space

Plots of  2 as a function of distance:-

*

 D  dv. 2  4 r 2 dr 2 (I) non-directional orbitals (s)

1s



2

r Narayana Junior Colleges

47


ATOMIC STRUCTURE

Plots of radial distribution functions 4pr2y2 for various orbitals (II) Directional orbitals a) Which depends only on one direction (P)


b) Which depends on two directions(d)

c) Which depends on three directions (f)

48



ATOMIC STRUCTURE

LEVEL - IV

8.

is subjected to a metal sheet having work function=12.8 eV. What will be the velocity of photo-electrons having maximum kinetic energy. (A) 0, no emission will occur (B) 4.352  106 m / s

SINGLE ANSWER QUESTIONS

2 0 (K.E) (A) 29e 2

(D)  K.E   29e 2 When the frequency of light incident on a metallic plate is doubled, the KE of the emitted photoelectron will be; (A) doubled (B) Halved (C) Increased but more than doubled of the previous KE (D)Remains unchanged Quantum number If 10–17 J of light energy is needed by the interior of human eye to see an object. The photons of green light (   550nm ) needed to see the object are (A)27 (B)28 (C)29 (D)30 A 1-kW radio transmitter operates at a frequency of 880 Hz. How many photons per second does it emit? (A) 1.71  1021 (B) 1.71  1030 23 (C)6.02  10 (D) 1.71  1033 Which of the following relates to photons both as wave motion and as a stream of particles? (A) Inference (B) E  mc 2 (C) Diffraction (D) E  h  (C)

4.

5.

6.

7.

4 0  K.E  

29e 2 (B) 2 K.E  0


9.

10.

(D) 8.72  106 m / s T he r at io of slopes of K maxvs. v and V0 vs. v curves in the photoelectric effect gives (v=frequency, kmax=maximum kinetic energy, V0=stopping potential): (A) charge of electron (B) Planck’s constant (C) work function (D) the ratio of Planck’s constant of electron ic charge Photoelectron emission is observed for three different metals A,B and C. The kinetic energy of the fastest photoelectrons versus frequency ‘ ’ is plotted for each metal. Which of the following graph shows the phenomenon correctly?

A B C K.E

3.

(C) 3.09  106 m / s

(A) Frequency (v)

A B C K.E

2.

Rutherford’s experiment, which established the nuclear model of the atom, used a beam of (A)  -particles, which impinged on a metal foil and got absorbed (B)  -rays, which impinged on a metal foil and ejected electrons (C) helium atoms, which impinged on a metal foil and got scattered (D) helium nuclei, which impinged on a metal foil and got scattered Given the abundances of isotopes 54Fe, 56Fe, 57 Fe are 5%, 90% and 5% respectively, the atomic mass of Fe is (A) 55.85 (B) 55.95 (C) 55.75 (D) 56.05 An  -particle approaches the target nucleus of copper(Z=29) in such a way that the value of impact parameter is zero. The distance of closest approach will be:


1.

0

Electromagnetic radiation having   310 A

(B)

Frequency (v)

49


ATOMIC STRUCTURE

13. A B K.E

The mass of an electron is m,charge is e and it is accelerated from rest through a potential difference of V volts. The velocity acquired by electron will be

2eV eV V eV (B) (C) (D) m m m 2m In two individual hydrogen atoms electrons move around the nucleus in circular orbits of radii R and 4R. The ratio of the time taken by them to complete one revolution is: (A)1 : 4 (B)4 : 1 (C)1 : 8 (D) 8 : 7 The difference in angular momentum associated with the electron in two successive orbits of hydrogen atoms is ( A)

C

14.

(C) Frequency

  15. A B C K.E

(D)

(A) 16.

11.

The given diagram indicates the energy levels of a certain atom. When system moves from 2E level to E level, a photon of wavelength  is emitted. The wavelength of photon produced during the transition 4E form to E level is is: 3


Frequency (v)

17.

18. 2E 4/3 E

19.

E 0

20.  3 4 (B) (C) (D) 3 3 4 3 Which of the following postulates does not belong to Bohr’s model of the atom?

(A) 12.

h 2 (B) The electron stationed in the orbit is stable (C) The path of an electron is circular (D) The change in the energy levels of electron is continuous

(A)Angular momentum is an integralmultiple of

50

21. 22.

h 

(B)

h 2

(C)

h h (D) ( n  1) 2 2

The ionization energy of a hydrogen atom in terms of Rydberg constant (RH) is given by the expression (A) RHhc (B)RHc (C) 2R Hhc (D)RHhcNA If the wavelength of series limit of Lyman series for He+ ion is x A0 , then what will be the wavelength of series limit of Balmer series for Li2+ ion? 9x 0 16x 0 5x 0 4x 0 A (B) A (C) A (D) A (A) 4 9 4 9 The potential energy of an electron in the hydrogen atom is -6.8 eV. Indicate in which excited state, the electron is present? (A) first (B) second (C) third (D) fourth What is the potential energy of an electron present in N-shell of the Be3+? (A) -3.4 eV (B) -6.8 eV (C) -13.6eV (D) -27.2 eV The distance between 4th and 3rd Bohr orbits of He+is: (A) 2.645x10-10m (B) 1.322x10-10m -10 (C) 1.851x10 m (D) 6.8x10-10m The ratio of velocity of the electron in the third and fifth orbit of Li2+ would be: (A) 3:5 (B) 5:3 (C) 25:9 (D) 9:25 If in Bohr’s model, for unielectronic atom, time period of revolution is represented as Tn,zwhere n represents shell no. and Z represents atomic number then the value of T1,2:T2,1 will be: (A) 8:1 (B) 1:8 (C) 1:1 (D) 1:32 Narayana Junior Colleges


ATOMIC STRUCTURE

24.

25.

The graph of E Vs . Z2 (keeping “n’ constant ) will be: Z2 E

(A)

(B)

E

Z2

E

30.

31.

32.

(D) Z2

Z2

33.

Potential energy of electron present in He+ is: e2 (A) 2 0 r

3e 2 (B) 4 0 r

2e 2 e 2 (C) (D) 4 0 r 4 0 r 2 The velocity of an e- in excited state of Hatom is 1.093  106 m / s . What is the circumference of this orbit? (A) 3.32  10 10 m (B) 6.64 1010 m (C) 13.30  1010 m (D) 13.28  108 m


The energy of a I, II and III energy levels 4E of a certain atom are E, and 2E 3 respectively. A photon of wavelength  is emitted during a transition from III to I. What will be the wavelength of emission for II to I?  (A) (B)  (C) 2 (D) 3 2 The angular momentum of an electron in hydrogen atom is proportional to 1 1 (A) r (B) (C) r 2 (D) r r The number of revolutions made by electron in Bohr’s 2nd orbit of hydrogen atom is (A) 6.55  1015 (B) 8.2  1014 (C) 1.64  1015 (D) 2.62  106 Ratio of frequency of revolution of electron in the 2nd excited state of He+ and 2nd state of hydrogen is:

1 27 32 27 (B) (C) (D) 27 32 54 2 A proton is accelerated from rest through a potential difference of ‘V’ volts has a wavelenght  associated with it. An alpha particle in order to have the same wavelength must be accelerated from rest through a potential difference of (A) V volt (B) 4V volt V (C) 2V volt (D) volt 8 If the 2nd excitation potential for a Hydrogen like atom in a sample is 108.9V. Then the series limit of the paschen series for this atom is: (A)

E

(C)

27.

29.

 13.6 2 Z eV/atom (Z=atomic number). n2

En 

26.

28.

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 inthe first exceited state of Li2+? (A) 3.4 eV (B) 10.2 eV (C) 30.6 eV (D) 6.8 eV The mass of a proton is 1836 times more than the mass of an electron. If a sub-atomic particle of mass (m!) 207 times the mass of electron is captured by the nucleus, then the first ionization potential of H: (A) decreases (B) increases (C) remains same (D) may be decrease or increase The energy of an electron moving in nth Bohr’s orbit of an element is given by


23.

34.

RH 32RH (A) RH (B) 2 (C) 2 (D) 32 R H 3 4 Let U1 be the frequency of the series limit of the Lyman series, u2 be the frequency of the first line of the Lyman series , and u3 be the frequency of the series limit of the Balmer series, then (A) u1  u 2  u 3 (B) u 2  u1  u 3

(C) u 3 

1  u1  u 3  2

(D) u1  u 2  u 3 51


36.

37.

38.

39.

40.

If the following matter waves travel with equal velocity, the longest wavelength is that of a/an. (A)electron (B)proton (C)neutron (D)particle If 1 and  2 denote the de-Broglie wavelength of two particles with same masses but charges in the ratio of 1 : 2 after they are accelerated from rest through the same potential difference, then (A) 1   2 (B) 1   2 (C) 1   2 (D) 1  2 . An electron in a H-like atom is in an excited state. It has a total energy of –3.4 eV, calculate the de-Broglie’s wavelength? (A)66.5Å (B)6.66Å (C)60.6Å (D)6.06Å The stationary Bohr’s orbit can be readily explained on the basis of wave nature of electron if it is assumed that (A) wave in any of the orbits is the stationary wave (B) the position of maxima and minima of wave does not change with time (C) the length of the circular orbit must be an integral multiple of the wavelength (D) wave in any of the orbit is not stationary wave Consider the following statements regarding Sommerfeld’s model. Select the correct statement/s. (A) Around the nucleus, some of the paths are elliptical and others are circular (B) When an electron revolves around the nucleus in a circullar path, the angle of rotation is changed. (C) Both, angle of rotation and distance from the nucleus, are changed when an electron revolves in an elliptical path. (D) All are correct The mass of a particle is 10 10 g and its radius is 2  104 cm . If its velocity is 106 cm sec 1 with 0.0001% uncertainty in measurement, the uncertainty in its position is : (A) 5.2  108 m (B) 5.2  10 7 m (C) 5.2  106 m (D) 5.2  109

52

41.

Which of the following graphs correctly represents the variation of particle momentum with associated de Broglie wavelength?

p

p

(A)

(B) 



p

p

(C)

(D) 

42.



de Broglie wavelengths of two particles A  1  and B are plotted against   : where V  V is the potential on the particles. Which of the following relations is correct about the mass of the particles?


35.

ATOMIC STRUCTURE

B A



1 V

(A) m A  m B 43.

44.

(B) m A  m B

(C) m A  m B (D) m A  m B A proton and an alpha particle are accelerated through the same potential difference. The ratio of the wavelenghts associated with the proton to that associated with the alpha particle is 1 (A) 4 (B) 2 (C) 8 (D) 8 The ratio of orbital angular momentum and spin angular momentum of an electron in ‘p’ orbital is (A)

3 2

(B)

3 2

(C)

2 2 3

(D)

2 3



ATOMIC STRUCTURE

46.

Probability of finding the electron ψ 2 of ‘s’ orbital does not depend upon (A) distance from the nucleus (r) (B) energy of ‘s’ orbital (C) principal quantum number (D) azimuthal quantum number The orbital angular momentum of an electron in 2s-orbital is

h (A) 4 47.

48.

49.

(B) zero

The subshell that arises after f is called the g subshell. How many electrons may occupy the g subshell? (A) 9 (B) 7 (C) 5 (D) 18 The quantum numbers of most energetic electron in Ar atom when it is in first excited state is (A)2, 1, 0,  1/2 (B)4, 1, 1,  ½ (C)4, 0, 0,  1/2 (D)4, 1, 0,  1/2 For a ‘d’ electron, the orbital angular

51.

(A) 6 (B) 2 (C)  (D) 2 The quantum numbers +1/2 and 1/ 2 for the electron spin represent (A) rotation of the electron in clockwise and anticlockwise direction respectively (B) rotation of the electron in anticlockwise and clockwise direction respectively (C) magnetic moment of the electron pointing up and down respectively (D) two quantum mechanical spin states which have no classical analogue The schrodinger wave equation for hydrogen atom is

53.

(A)

a0 2

(B) 2a 0


(C)

2a 0

(D)

a0 2

6

h 

(C)

3

h h (D) 15  

If 0.53 A0 is Bohr’s radius for the first orbit. It suggest in the light of the wave mechanical model that it reaches at the distance of 0.53 A0 (B) only  2 goes on increasing, 4r 2 dr remains constant till it reaches at the distance of 0.53 A0 (C)  2 goes on increasing , 4r 2 dr goes on decreasing till it reaches at the distance of 0.53 A0 (D) only 4r 2 dr goes on increasing ,  2 remains constant till it reaches at the distance of 0.53 A0

54.

55.

56.

57.

where a 0 is Bohr’s radius. If the radial node in 2s be at r 0, then r0 would be equal to:

h (B) 

(A) the product of  2 and 4r 2dr increase till

3

1  1  2 r   a 0  2s    2  e a0  4 2  a 0  

For a ‘f’ electron the orbital angular momentum is (a) 12

h 2h (C) (D) 2 2

h   momentum is     2  

50.

52.


45.

Magnetic moments of V(Z = 23), Cr(Z = 24), Mn(Z = 25) are x, y, z. Hence: (A) x = y = z (B) x < y < z (C) x < z < y (D) z < y < x The value of the magnetic moment of a particular ion is 2.83 Bohr magneton. The ion is (A) Fe2+ (B) Ni2+ (C) Mn2+ (D) Co3+ If the nitrogen atom has electronic configuration 1s7, it would have energy lower than that of the normal ground state configuration 1s 2 2s 2 2p 3 because the electrons would be closer to the nucleus. Yet 1s7 is not observed because it violates (A) Heisenberg uncertainty principle (B) Hund’s rule (C) Pauli exclusion principle (D) Bohr postulate of stationary orbits If the subsidiary quantum number of a subenergy level is 4, the maximum and minimum values of the spin multiplicites are : (A) 9,1 (B) 10,1 (C) 10,2 (D) 4,-4

53


58.

ATOMIC STRUCTURE

The orbital diagram in which both the Pauli’s exclusion principle and Hund’s rule are violated,is: (A)

(B)

(C)

(D)

64.

MULTIPLE ANSWER QUESTIONS

60.

61.

62.

When alpha particles are sent through a thin metal foil, most of them go straight through the foil because: (A) alpha particles are much heavier than electrons (B) alpha particles are positively charged (C) most part of the atom is empty space (D) alpha particle move with high velocity Many elements have non-integral atomic masses because (A) they have isotopes (B) their isotopes have non-integral masses (C) their isotopes have different masses (D) the cosstituents, neutrons, protons and electrons combine to give fractional masses Which statement about cathode rays is correct? (A) They travel in straight lines towards cathode (B) They produce fluorescent discharge through the walls of the tube (C) They produce heating effect (D) They can affect photographic plate Which statements concerning Bohr’s model are true? (A) Predicts that probability of electron near nucleus is more (B) Angular momentum of electron in H-atom =

65.


59.

66.

than that of partilce B  m A  m B  ?

67.

68.

nh 2

63.

54

(C) Introduces the idea of stationary states (D) Explain line spectrum of hydrogen In Rutherford’s gold foil experiment, the scattering of  -particles takes place. In this process: (A) coulombic force is involved (B) nuclear force is involved (C) path of  -particle is parabolic (D) path of  -particle is hyperbolic

Choose the correct statement(s) regarding the photo-electric effect. (A) No electrons are ejected, regardless of the intensity of the radiation, unless the frequency exceeds a thershold value characteristic of the metal (B) The kinetic energy of the ejected electrons varies linearly with the frequency of the incident radiation and its intensity (C) Even at low intensities, electrons are ejected immediately if the frequency is above the threshold value (D) An intense and a weak beam of monochromatic radiations differ in having number of photons and not in the energy of photons The energy of an electron in the first Bohr orbit of H atom is 13.6eV . The possible energy value(s) of the excited state(s) for electrons in Bohr orbits of hydrogen is/are (A) 3.4 eV (B) 4.2 eV (C) 6.8eV (D) 6.8eV In which of the following conditions the de Broglie wavelength of particle A will be less

69.

(A) Linear momentum of these particles are same (B) Move with same speed (C) Move with same kinetic energy (D) have fallen through same height Which of the following quantum numbers is/ are not allowed? (A) n  3, l  2, m  0 (B) n  2, l  2, m  1 (C) n  3, l  0, m  1 (D) n  5, l  2, m  1 Which represent a possible arrangement ? n m s  (A) 3 2 –2  1/2 (B) 4 0 0  1/2 (C) 3 2 –3  1/2 (D) 5 3 0  1/2 T he pr obability of finding the electr on in p xorbital is (A) maximum on two opposite sides of the nucleus along x-axis (B) zero at the nucleus (C) same on all the sides around the nucleus (D) zero on the z-axis Narayana Junior Colleges


ATOMIC STRUCTURE

71.

72.

73.

74.

75.

‘g’ orbital is possible if (A) n = 5, l  4 (B) It will have 18 electrons (C) Sublevel will have 9 orbitals (D) It will have 22 electrons Choose the correct statement(s): (A) For a particular orbital in hydrogen atom, the wave function may have negative value (B) Radial probability distribution function may have zero value but can never have negative value (C) 3d x 2  y 2 orbital has two angular nodes and one radial node. (D) yz and xz planes are nodal planes for d xy orbital Choose the correct statement(s): (A) Heisenberg’s principle is applicable to stationary e (B) Pauli’s exclusion principle is not applicable to photons (C) For an e , the product of velocity and principal quantum number will be independent of principal quantum bumber (D) Quantum number l and m determine the value of angular wave function Select the correct statements about the wave function  . (A)  must be real (B)  must be single valued, continuous (C)  has no physical significance (D)  2 gives the probability of finding the electrons Which of the following is/are correct energy order for H-atom? (A) 1s < 2s < 2p < 3s < 3p (B) 1s < 2s = 2p < 3s = 3p (C) 1s < 2p < 3d < 4s (D) 1s < 2s < 4s < 3d Ground state electronic configuration of nitrogen atom can be represented by

(A) (B) (C) (D) Narayana Junior Colleges

ASSERTION & REASON QUESTIONS 76.

77. Narayana Junior Colleges

70.

78.

Statement- 1 : The kinetic energy of photoelectrons increases with increase in frequency of incident light where    0 . Statement - 2: Whenever intensity of light is increased the number of photo-electron ejected always increases. (A) If both the statement are TRUE and STATEMENT -2 is the correct explanation of STATEMENT -1 (B) If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1 (C) If STATEMENT-1 is TRUE and STATEMENT -2 is FALSE (D) If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE Statement -1: Half -filled and fully-filled degenerate orbitals are more stable Statement - 2: Extra stability is due to the symmetrical distribution of electrons and exchange energy (A) If both the statement are TRUE and STATEMENT -2 is the correct explanation of STATEMENT -1 (B) If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1 (C) If STATEMENT-1 is TRUE and STATEMENT -2 is FALSE (D) If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE Statement -1: The ground state configuration of Cr is 3d 5 4s1 Statement - 2:A set of half-filled orbitals containing one electrons each with their spin parallel provides extra stability. (A) If both the statement are TRUE and STATEMENT -2 is the correct explanation of STATEMENT -1 (B) If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1 (C) If STATEMENT-1 is TRUE and STATEMENT -2 is FALSE (D) If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE 55


79.

ATOMIC STRUCTURE

STATEMENT-1 For hydrogen orbital energy increases as 1s < 2s < 2p < 3s < 3p < 3d < 4s < 4p … STATEMENT-2 The orbital with lower (n +  ) value has lesser energy and hence filled up first. (A) If both the statement are TRUE and STATEMENT -2 is the correct explanation of STATEMENT -1 (B) If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1 (C) If STATEMENT-1 is TRUE and STATEMENT -2 is FALSE (D) If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE

80.

81.

PASSAGE 1 For a single electron atom or ion the wave number of radiation emitted during the transition of electron from a higher energy state (n = n2) to a lower energy state (n = n1) is given by the expression:  1 1 1     R H .z 2  2  2   n2   n1

where R H 

… (1)


COMPREHENSIONS QUESTIONS

PASSAGE 2 A german physicist gave a principle about the uncertainties in simultaneous measurement of position and momentum of small particles. According to that physicist. It is impossible to measure simultaneously the position and momentum of small particle with absolute accuracy or certainty. If an attempt is made to measure any one of these two quantities with higher accuracy , the other becomes less accurate. The product of the uncertainty in position

 x 

m.m m  m where m ' = mass of nucleus For Lyman series : n1 = 1 (fixed for all the lines) while n2 = 2, 3, 4, … for successive lines i.e. 1st, 2nd, 3rd … lines, respectively. For Balmer series : n1 = 2 (fixed for all the lines) while n2 = 3, 4, 5 … for successive lines.

h 4 If uncertainty in position is twice the uncertainty in momentum, then uncertainty in velocity is

 x  p  

82.

(A)



56

and uncertainty momentum  p  is

always constant and is equal to or greater than h / 4 , where h is Planck’s constant i.e.

22 m k 2e 4 = Rydberg constant for h 3c

H-atom where the terms have their usual meanings. Considering the nuclear motion, the most accurate expression would have been to replace mass of electron (m) by the reduced mass (  ) in the above expression, defined as

The ratio of the wave numbers for the highest energy transition of e– in Lyman and Balmer series of H-atom is: (A)4 : 1 (B)6 : 1 (C)9 : 1 (D)3 : 1 If proton in H-nucleus be replaced by positron having the same mass as that of electron but same charge as that of proton, then considering the nuclear motion, the wavenumber of the lowest energy transition of He+ ion in Lyman series will be equal to (A)2 RH (B)3 RH (C)4 RH (D)RH

h 

(B)

1 h 2m 

1 h 1 h (D) 2m 2 2m  The uncertainty in position of an electron

(C) 83.

m  9.1 10

28



gm moving with a velocity

3  10 4 cm / s accurate upto 0.001% will be (A) 3.84 cm (B)1.92 cm (C)7.68 cm (D)5.76 cm Narayana Junior Colleges


ATOMIC STRUCTURE

88.

If uncertainty in the position of an electron is zero , the uncertainty in its momentum would be (A) zero (B)  h / 4 (C)  h / 4 (D)Infinite

PASSAGE 3: In the Rutherford’s experiment, -particles were bombarded towards the copper atoms so as to arrive a distance of 10-13 metre from the nucleus of copper and then getting either deflected or traversing back. The particles did not move further closer 86. The velocity of the -particles must be (A) 8.32  108 cm / sec (B) 6.32 108 cm / sec (C) 6.32  108 m / sec (D) 6.32  108 km / sec 87. From Rutherford’s -particle scattering, it can be concluded that   1 (A) N sin    (B) N  sin 4   2 1  (C) N (D) N  sin 4 sin ( / 2) 2 88.It can also be concluded that the electrostatic potential energy is equal to 1 q1q2 1 5Ze 2 (A) 4 r (B) 4 0 mv 2 0

Ze 2 r PASSAGE -4 (C)

The orbital angular momentum of electron    1 makes an angle of 450 from Z-axis. The LZ of electron will be h  h   h   h  (A) 2  2   (B) 0  2   (C) (D) 3       2  2  89. An orbital has n=5 and its  value is the maximum possible. The orbital angular momentum of the electron in this orbital will be h h (A) 2 (B) 6 2 2 h  h  (C) 12 (D) 20   2  2  PASSAGE 5 The sum of spins of all the electrons is the total spin(S) and (2S + 1) is called spin multiplicity of the electronic configuration. Hund’s rule defines the ground state configuration of electrons in degenerate orbitals i.e., orbitals within the same sub-shell which have the same values of n and l , states that in degenerate orbitals pairing of electrons does not occur unless and until all such orbitals are filled singly with their parallel spin. A spinning electron behaves as though it were a tiny bar magnet with poles lying on the axis of spin. The magnetic moment of any atom, ion or molecule due to spin called spin-only magnetic moment (ms) is given by the formula.

(D) mv 2 90.

h L      1 2 On the other hand, m determines Zcomponent of orbital angular momentum as  h  Lz  m    2  Hund’s rule states that in degenerate orbitals electron s do not pair up unless and until each such orbital has got an electron with parallel spins. Besides orbital motion, an electron also possess spin -motion. Spin may be clockwise and anti-clockwise. Both thes spin motions are called two spin states of electron characterised by spin.

Q.N (s): s  


84.

91.

 s  n n  2  B.M where n = number of unpaired electron(s) The spin-only magnetic moment of Cr3+ ? (A) 3 B.M. (B) 8 B.M. (C) 15 B.M. (D)Zero The spin-multiplicity of Fe3+ (Ec=[Ar]3d5) in its ground state (A)6 (B)2 (C)3 (D)4 INTEGER TYPE

92. 93.

Spin multiplicity of Nitrogen atom is The work function    of some metals is listed below. The number of metals which will show photoelectric effect when light of 300 nm wavelenght falls on the metal is (2011)

1 1 and s   respectively.. 2 2


57


ATOMIC STRUCTURE

Difference between nth and (n + 1)th Bohr’s radius of H atom is equal to it’s (n – 1)th Bohr’s radius. The value of n is 95. A single electron system has ionisation energy 11180 KJ mole-1. The number of protons in the nucleus of the system is ... 96. The number of spectral lines produced when an electron jumps from 5th orbit to 2nd orbit in the hydrogen atom is. 97. In an collection of H-atoms , all the electrons jump form n=5 to ground level finally (directly or indirectly), without emitting any line in Balmer series. The number of possible different radiations is: 98. In a single isolated atom an electron make transition from 5th excited state to 2nd state then maximum number of different types of photons observed is 99. The number of waves made by a Bohr electron in Hydrogen atom in one complete revolution in the 3rd orbit is. 100. The minimum number of waves made by an electron moving in an orbit having maximum magnetic quantum number +3 is. 101. The wave function of an orbital is 94.

number of unpaired electrons in M x  is. 105. How many d – electrons in Cu  ( At.No  29)  1  

can have the spin quantum    ? 2 106. The maximum number of electrons can have pricipal quantum number n = 3 and spin quantum number Mz  

58

1 is (2011) 2

MATRIX MATCHING TYPE 108. Column I (A)  310

Column II (p) 5 f

(B)  120

(q) 3 px or 3 p y

(C)  530

(r) 3 pz

(D)  311

(s) impossible

109. Column I (A) Thomson model of atom


represented as  420 . The azimuthal quantum number of that orbital is 102. The radial distribution curve of the orbital with double dumbbell shape in the 4 th principle shell consists of ‘n’ nodes, n is 103. A compound of vanadium possesses a magnetic moment of 1.73 BM. the oxidation state of vanadium in this compounds is: 104. Magnetic moment of M x  is 24BM . The

107. One mole of photons, each of frequency 250 sec-1 would have approximately a total energy in ergs

Column II (p) Electrons are present in extra nuclear region (B) Rutherford model (q) Atom is electrically of atom neutral (C) Bohr model of atom (r) Positive charge is accumulated in the nucleus (D) Sommerfeld model (s) Uniform sphere of of atom positive charge with embeded electrons 110. Column I Column II (A) Radial function ( r ) (p) Principle Q.No. (B) Angular function ( ) (q) Azimuthal Q.No. (C) Angular function ( ) (r) Magnetic Q.No (D) Quantized angular momentum (s) Spin Q.No (t) Shape of orbital 111. Column- I Column- II th (A) Radius of n orbit (p) Inversely proportional to z th (B) Energy of n orbit (q) Integralmultiple of `

h 2

(C) Velocity of electron (r) Proportional to x1010 m in the nth orbit (D) Angular momentum (s) Inversely of electron proportional to n (t) Inversely proportional to n 2 Narayana Junior Colleges


ATOMIC STRUCTURE

112. Column- I Column- II (Phenomenon related (Character of the to electron) electron) (A) Working of electron (p) Wave nature microscope (B) Photoelectric effect (q) Particle nature (C) Diffraction (r) Particle nature dominates the wave nature (D) Scintillation (s) Wave same dominates the particle nature. 113. Column I Column II (A) 2p orbital (p) Number of spherical nodes =0 (B) 3d orbital (q) Number of nodal plane =0 (C) 2s orbital (r) Orbital angular momentum number =0 (D) 4f orbital (s) Azimuthal quantum number=0 114. Column I Column II (P) s  s  1

h 2

momentum of an electron (B) Angular momentum (Q)

n  n  2

of an electron in an orbit (C) Spin angular

(R)

3.

4.

1 h 2 2

(C) 5.

6.

nh 2

momentum of an electron (D) Magnetic moment

(S)

    1

h 2

7.

of atom PREVIOUS YEARS IIT JEE QUESTIONS SINGLE CORRECT ANSWER TYPE 1.

Which of the following relates to photons both as wave motion and as a stream of particles? (A) Inference (B) E  mc 2 (C) Diffraction (D) E  h 


Which of the following does not characterise X-rays? (A) The radiation can ionize gases (B) It causes ZnS to fluoresce (C) Deflected by electric and magnetic fields (D) Have wavelengths shorter than ultraviolet rays A 3p orbital has (A) two non-spherical nodes (B) two spherical nodes (C) one spherical and one non-spherical node (D) one spherical and two non-spherical nodes The orbital angular momentum of an electron in 2s orbital is (A)  .


(A) Orbital angular

2.

8.

h 2

(B) zero (D) 2.

h 2

The first use of quantum theory to explain the structure of atom was made by (A) Heisenberg (B) Bohr (C) Planck (D) Einstein For a d-electron, the orbital angular momentum is (A) 6  h / 2 

(B) 2  h / 2 

(C)  h / 2 

(D) 2  h / 2 

The electrons, identified by quantum numbers n and l , (i)n = 4, l = 1, (ii) n = 4, l =0,(iii)n=3, l = 2, and (iv) n = 3, l = 1 can be placed in order of increasing energy, from the lowest to highest, as (1999) (A) (iv) < (ii) < (iii) < (i) (B) (ii) < (iv) < (i) < (iii) (C) (i) < (iii) < (ii) < (iv) (D) (iii) < (i) < (iv) < (ii) The number of nodal planes in a px orbital is (2000) (A) one (B) two (C) three (D) zero 59


9.

ATOMIC STRUCTURE

The radius of which of the following orbits is same as that of the first Bohr’s orbit of hydrogen atom? (A) He+ (n = 2) (B) Li2+ (n = 2) (C) Li2+( n = 3) (D) Be3+ (n = 2)

83) B 84) D 85) D 86) B 87) C 88) C 89) D 90) C 91) A 92) 4 93) 4 94) 4 95) 4 96) 6 97) 7 98) 4 99) 3 100) 4 101) 2 102) 1 103) 4 104) 4 105) 5 106) 9 107) 1 108) A-r; B-s; C-p; D-q 109) A-q,s; B-p,q,r; C-p,q,r; D-p,q,r 110) A-p,q; B-q,t; C-r,t; D-q,s 111) A-p,r; B-r,t; C-s; D-q 112) A-p,s; B-q,r; C-p; D-q 113) A-p; B-p; C-q,r,s; D-p 114) A-s; B-r; C-p; D-q

MULTIPLE ANSWER TYPE Which of the following statement(s) is/are correct? (1998) (A) The electronic configuration of Cr is [Ar]3d54s1. (Atomic number of Cr = 24) (B) The magnetic quantum number may have a negative value. (C) In silver atom, 23 electrons have a spin of one type and 24 of the opposite type. (Atomic number of Ag = 47) (D) The oxidation state of nitrogen in HN 3 is 3

LEVEL - IV - KEY 1) D 5) B 9) A 13) A 17) B 21) B 25) B 29) A 33) A 37) B 41) D 45) D 49) A 53) A 57) B 60) A,C 63) A,D 66) B,D 69) A,B,D 72)B,C,D 75) A,D 79) D

60

2) B 3) B 6) D 7) D 10) C 11) D 14) C 15) B 18) A 19) D 22) D 23) C 26) C 27) C 30) B 31) A 34) A 35) A 38) C 39) D 42) B 43) C 46) B 47) D 50) D 51) B 54) C 55) B 58) D 59) A,C 61) B,C,D 62) B,C,D 64) A,C,D 65) A 67) B,C 68) A,B,D 70) A,B,C 71) A,B,D 73) A,B,C,D 74) B,C 76) B 77) A 80) A 81) B

4) C 8) C 12) D 16) A 20) C 24) B 28) D 32) D 36) C 40) A 44) C 48) C 52) C 56) C

PREVIOUS IIT JEE QUESTIONS

KEY Narayana Junior Colleges

10.

1.(D) 5.(B) 9.(D)

4.(B) 8.(A)

LEVEL - IV - HINTS 1. 2. 3. 4.

5. 6. 7. 8. 9.

78) A 82) D

2.(C) 3.(C) 6.(A) 7. (A) 10.(A), (B), (C)

It is Rutherford  - scatering experiment PM  P M  P M Avg: At wt  1 1 2 2 3 3 100 1 QQ r0  . 1 2 4 0  KE  h =h 0 +KE , threshold frequency of metal is constant hc E  n.  hc E  n. (1kW =1000J)  Plank’s quantum theory hc 1    mv 2  2 KE  h  h 0 eV0  h  h 0

h ratio of slopes =  h / e   e Narayana Junior Colleges


ATOMIC STRUCTURE

10.

11.

12. 13.

Different metals have different work function ie different threshold frequencies 2E  E  E 

4 E E   ? 3 3 Bohr’s concept

T1 n13  Where n1  1, n 2  2 T2 n 32

15.

nh mvr  2

26.

PE  

27.

Vn  2.18  108 / n cm / sec Where n=2

28.

Apply E 

29.

mvr 

30.

17. 18. 19.

PE  2  T.E 

  13  Z 2   2  2  n 

rn ,z

21.

Tn,z 

n3 z2

22.

Tn,z 

n3 z2

31.

f n,z 

32.



Z n2

I.E

24.

I.E  R H .ch

RH 

2 2 4e 4 .K 2 Ch 3

m1.m Where   reduced mass  1 m m Narayana Junior Colleges

2rn,z

z2 n3

h 2mq.v

  p x

33.

m p. 1 .V 4m p .  2  .x

v 8

 1 1  E  13.6  z 2  2  2   n1 n 2  z=3  1   R H .z 2  2   3 

2

23.

Vn,z

z2 6.56  10  3 n

n2  0.529  Z

20.

f n,z 

15


1 1 1     2  2  R H .z 2   n1 n 2  PE=2(T.E)

nh 2

 angular momentum of e   r

    R H ch 



hc 

rn n 2

IE   E1

 R H ch

1 ze2 . 40 r

Circum ference = 2rn

1 n  vn  cons tan t

14.

y   mx plot

 New wave length = 3

v

Tn 3 

16.

hc 

25.

 RH

34. 

1

3

3 2

2 1

61


h mv

35.



36.

h  2.m.q.v

37.

n=2

38. 39.

2 rn  n n sommerfed’s concept h x  where v  1012 4m.v h    .p  constant graph p

40. 41.

ATOMIC STRUCTURE

2rn  n n   n 

2rn n

54.

  n  n  2

55.

  n  n  2  Where n=2

56. 57. 58. 59.

Pauli exclusion principle concept Spin multiplicity =2s+1 filling rules concept Ruther ford’s   ray experiment concept

60.

At.wt 

61. 62. 63. 64.

Cathode rays properties Bohr’s concept Ruther ford’s  -ray experiment concept Photo electric effect concept

70.

13.6 ev / atom n2 Quantumm numbers concept Quantum Numbers concept Px-orbital has dumb bell shape and is situated along x-axis Orbital concept71. Orbital concept

72.



h 2m.q.v

65.

43.

h  2m.q.v

66. 68. 69.

44.

Orbital angular momentum= L      1 .

h 2 2 For s orbital,  depends only on ‘n’ quantums no h L      1 . 2 g subshell has 9 orbitals quantum numbers concept h L      1 . 2 Spin quantum number concept At node  2  0 hence   0

Spin angular momentum= s  s  1 . 45.

46. 47. 48. 49. 50. 51.


42.



h 2

P1M1  P2 M 2  ... p1  p 2  ....

En  

1 n

  r,  ,    R n,l  r  . l,m    . m   

73. 74. 75. 76. 77. 78. 79.

Properties of  In H-atom, all are degenerate orbitals in a given main shell Hund’s maximum multiplicity rule h  h 0  KE Electronic configuratio concept Electronic configuration concept For H-atom, all are degenerate orbitlas in a given main shell

 r  2a 0

52.

53.

h L      1 . 2

80.

 1 1   R H .z 2  2  2   n1 n 2 

81.

 1 1   R H .z 2  2  2  =3R H  n1 n 2 

82.

x  2 p

x.v 

62

h 4m Narayana Junior Colleges


ATOMIC STRUCTURE

83.

x 

102. no. of radial nodes  n    1

h 4m.v

103.  s  n  n  2  Where n=1

h 4 85. CONCEPTUAL 86.At the distance of closest approach

84.

104.  s  n  n  2  Where n=4

x. p 

v2 

105. configuration is  Ar  3d10 106. n = 3, l = 0,1,2

2 Ze2e 1  4 0 r0 m hence, substituting the value of

for l = 0

1 1 m = 0 mz   ,  2 2

l=1

1 1 m = -1 mz   ,  2 2

l=1

1 1 m = 0 mz   ,  2 2

l=1

1 1 m = 1 mz   ,  2 2

m = 4, Z = 29, e = 1.66 x 10-19, ro = 10-13m.

87.CONCEPTUAL

89.

L z  L cos  L      1 .

h 2

90.

s  n  n  2

91. 92. 93.

Spin multiplicity =2S+1 Spin multiplicity = 2S + 1 Incident radiation energy is 4.13 ev Condition : E  E 0

94.

0

R n  0.529  n 2 A

52  4 2  32

1312 Z2 2 n

95.

I .E. 

96.

no. of spectral lines =


88.

n  n  1 2

n  n 2  n 1 5 4 3

97.

2 1

98.

In a single isolated atom, no .of spectral lines formed=n2-n1.

99. no. of waves in n th orbit  n 100.CONCEPTUAL 101

for l = 2

1 1 m = -2 mz   ,  2 2

for l = 2

1 1 m = -1 mz   ,  2 2

for l = 2

1 1 m = 0 mz   ,  2 2

for l = 2

1 1 m = +1 mz   ,  2 2

for l = 2

1 1 m = +2 mz   ,  2 2

OR n=3 number of electors  2n 2  2  32  18 1  18  electons with  ms      9 2 2  107. Total energy = N0hv 108. Based on quantam mechanics 109. Different atomic models concept 110. Based upon orientation of electrons

111. rn,z 

n2 z

E n,z  

z2 n2

 n,l,m is the wave function of given orbital


63


ATOMIC STRUCTURE

Hence, the plane yz is called nodal plane as it has zero electron density in it.

z mvr  n n 112. Dual nature of light concept 113. no. of radial nodes  n    1 no.of nodal palnes =  h L      1 . 2 Vn,z 

h 114. L      1 . 2

z

x

nh mvr  2

9.

o

rH  0.529 A

for Be 2+ ,

Magnetic moment  s  n  n  2 

E  mc 2 as particles.

2.

X-rays being neutral are not deflectd by magnetic and electric fields.

3.

Number of radial nodes =  n  l  1 Fo r 3porbital n = 3, l  1 for p-orbital Number of spherical or radial nodes = 3 - 1- 1 =1 Number of angular nodes = l For 3p, l = 1 for p-orbital, angular node = 1 It has one spherical and one non-spherical node.

4.

  l  l  1

5.

orbital angular momentum   0.0 Bohr made use of quantum theory and gave the structure of atom.

6.

  l  l  1   23

7. 8.

64

10.


c  h relates for wave motion. 

n = 2, Z = 4

=

o 0.529  n 2 0.529  2 2   0.529A Z 4

PREVIOUS YEARS IIT JEE QUESTIONS HINTS E  h

0.529n 2 o A Radius of an orbit = Z

In case of hydrogen Ist orbit n = 1, Z = 1,

h Spin angular momentum = s  s  1 . 2

1.

y

(A) Electronic configuration of Cr (Ar) 3d54s1. (B) Magnetic quantum can have a negative value as m = l to l . (C) In Ag[Kr]4d105s1 23 electrons have spin of one type and 24 of the opposite type. 24 electrons have spin of other type due to 5s1. ***

h for s-orbital l  0 . Hence, 2

h ; For d-orbital l  2 ; 2

h h  6 2 2

Aufbau’s principle Nodal plane is that plane at which probability of finding electron density is zero. In case of px the dumb-bell shape orbital has two lobes on x-axis. Narayana Junior Colleges

2a.atomic Structure( 41-64 )

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