P HYSICS F UNDAMENTAL F OR IIT-J EE
Atomic Structure, X-Ray & Radio Activity KEY CONCEPTS & PROBLEM SOLVING STRATEGY Atomic Structure : According to Neil Bohr's hypothesis is the angular momentum of an electron is quantised.
The maximum number of electrons that can be accommodated in an orbit is 2n2. X-rays :
h h mvr = n or L = n 2π 2π
When fast moving electron strikes a hard metal, X-rays are produced. When the number of electrons striking the target metal increases, the intensity of Xrays increases. When the accelerating voltage/kinetic energy of electron increases λmin decreases. X-rays have the following properties :
2πr = nλ h z –1 c = × ms 2πmr 137 n
h2 rn = 2 2 4π mke ke 2 fn = hr
K.E. =
n2 n2 1 = 0.529 Å where k = Z Z 4 πε 0
(a) Radiations of short wavelength (0.01 Å – 10Å); high pentrating power; having a speed of 3 × 108 m/s in vacuum.
15 × 1 = 6.58 × 10 Hz n n
Intensity
vn = Zn
1 ke 2 Z ke 2 − ke 2 ; P.E. = × Z; T.E. = – ×Z 2 r r 2r − 13.6 Z 2
ev/atom where –13.6 n2 = Ionisation energy + P.E. ⇒ +T.E. = = – K.E. 2 Note : If dielectric medium is present then εr has to be taken into consideration. T.E. =
λmin
(c)
1 1 = R(Z – b)2 1 − 2 λ n
(d) Moseley law
1 p mv 1 = RZ 2 − 2 = = h h n1 n 2
n=2
hc hc 12400 = = Å eV K.E V
b = 1 for k-line transfer of electron
2
n=3
λ
(b) λmin =
me 4 z 2 1 1 1 v =v= = − 2 2 3 2 c λ 8ε 0 h c n1 n 2
n=∞ n=7 n=6 n=5 n=4
Continuous spectrum (Varies & depends on accelerating voltage) Characteristic spectrum Kα (fixed for a target material) Kβ Lγ Lβ Lα
R = R0A
1/3
ν = a(z – b)
where R0 = 1.2 × 10–15 m
R = radius of nucleus of mass number A. * Nucleus density is of the order of 1017 kg/m3
Kδ
Lγ Lβ
Kγ Kβ
Lα
Paschen (I.R.)
Isomers are nuclides which have identical atomic number and mass number but differ in their energy states.
–0.85 eV Pfund Brackett (I.R.) (I.R.) –1.5 eV
∆mc 2 Nuclear binding energy = A Nucleon
–3.4 eV
Balmer (Visible) Limiting line of Lyman series
n=1
where ∆m = mass defect –13.6 eV
=
Lyman Series (U.V. rays)
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[ Zm p + (A − Z)m n − M ]c 2 A
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* The binding energy per nucleon is small for small nuclei.
(f) In a nuclear fusion reaction small nuclei fuse to give big nuclei whereas in a nuclear fusion reaction a big nuclei breaks down.
* For 2 < A < 20, there are well defined maxima which indicate that these nuclei are more stable.
Thermal neutrons produce fission in fissile nuclei. Fast moving neutrons, when collide with atoms of comparable masses, transfer their kinetic energy to colliding particle and slow down.
* For 30 < A < 120 the average B.E./A is 8.5 MeV / nucleon with a peak value of 8.8 MeV for Iron. * For A > 120, there is a gradual decreases in B.E./nucleon. * More the B.E./A, more is the stability. Radioactivity :
η=
(a) N = N0e–λt
out put In put
dN −dN = λN where = activity level dt dt n
Solved Examples
t
1 1 T1 / 2 (c) N = N0 = N0 2 2
1.
n
1 ⇒ A = A0 where A = activity level 2
(d) T1/2 = (e) τ =
=
The energy of an excited hydrogen atom is –3.4 eV. Calculate the angular momentum of the electron according to Bohr theory.
Sol. The energy of the electron in the nth orbit is
0.693 λ
En = –
1 λ
Here,
(f) τ = 1.4 T1/2 (g) t =
E nhν nhc = = t t λt
Power, P =
β particles are electrons emitted from the nucleus. (n → p + β)
(b)
–
or
N A 2.303 2.303 log10 0 = log10 0 λ N λ A
13.6 n2
13.6 n2
eV
= –3.4
n=2
Angular momentum =
2 × 6.63 × 10 −34 nh = 2π 2 × 3.14
= 2.11 × 10–34 Js.
m 2.303 log 0 λ m
2.
(h) If a radioactive element decays by simultaneous −dN = λ1N + λ2N emission of two particle then dt The following parameters remain conserved during a nuclear reaction
The wavelength of the first member of the Balmer series in the hydrogen spectrum is 6563 Å. Calculate the wavelength of the first member of the Lyman series.
Sol. For the first member of the Balmer series 5R 1 1 1 = R 2 − 2 = λ 36 2 3
(a) linear momentum (b) Angular momentum
...(1)
For the first member of the Lyman series
(c) Number of nucleons
1 1 3R 1 = R 2 − 2 = λ´ 4 2 1
(d) Charge (e) The energy released in a nuclear reaction
...(2)
Dividing Eq. (1) by Eq. (2)
X+P→Y+Z+Q
λ´ 5× 4 5 = = λ 36 × 3 27
Q = [mx + mp) – (my + mz)]c2 = ∆m × c2 Q = ∆m × 931 MeV or XtraEdge for IIT-JEE
v ∆λ = λ c
According to Doppler's effect of light
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λ´ =
5 5 λ= × 6563 = 1215 Å 27 27
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3.
Hydrogen atom in its ground state is excited by means of a monochromatic radiation of wavelength 970.6 Å. How many different wavelengths are possible in the resulting emission spectrum ? Find the longest wavelength amongst these.
Sol. Initial kinetic energy of the electron = 50.0 keV
Energy of the photon produced in the first collision, E1 = 50.0 – 25.0 = 25.0 keV Wavelength of this photon
Sol. Energy the radiation quantum
E = hv =
λ1 =
hc 6.6 × 10 −34 × 3 × 108 = λ 970.6 × 10 −10 × 1.6 × 10 −19
= 0.99 × 10–10 m = 0.99 Å
= 12.75 eV
Kinetic energy of the electron after third collision = 0
Energy of the excited sate
Energy of the photon produced in the third collision ,
En = – 13.6 + 12.75 = – 0.85 eV Now, we know that En = – or n2 = –
6.6 × 10 −34 × 3 × 108 hc = E1 1.6 × 10 −19 × 12.5 × 10 3
E3 = 12.5 – 0 = 12.5 keV
13.6
This is same as E2. Therefore, wavelength of this photon, λ3 = λ2 = 0.99 Å
n2
−13.6 13.6 = = 16 En − 0.85
5.
or n = 4 The number of possible transition in going to the ground state and hence the number of different wavelengths in the spectrum will be six as shown in the figure. n 4 3
In an experiment on two radioactive isotopes of an elements (which do not decay into each other), their mass ratio at a given instant was found to be 3. The rapidly decaying isotopes has larger mass and an activity of 1.0 µCi initially. The half lives of the two isotopes are known to be 12 hours and 16 hours. What would be the activity of each isotope and their mass ratio after two days ?
Sol. We have, after two days, i.e., 48 hours, 4
1 N1 = N10 = N10 /16 2
2
3
1 N2 = N 02 = N 02 /8 2 1
Mass ratio =
The longest wavelength corresponds to minimum energy difference, i.e., for the transition 4 → 3. Now
E3 = –
13.6 32
Now,
= – 1.51 eV
λmax =
A1 = λ1N1 = λ1 N10 /16 = A10 /16 = (1/16)µCi A2 = λ2N2 = λ2 N 02 /8
6.6 × 10 −34 × 3 × 108 (1.51 − 0.85) × 1.6 × 10 −19
= 18.75 × 10–7m = 18750 Å 4.
X-rays are produced in an X-ray tube by electrons accelerated through a potential difference of 50.0 kV. An electron makes three collisions in the target before coming to rest and loses half its kinetic energy in each of the first two collisions. Determine the wavelengths of the resulting photons. Neglect the recoil of the heavy target atoms.
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A10 = λ1 N10 = 1.0 µCi
After two days,
hc = E4 – E3 λ max
or
N0 8 N1 3 3× 8 = 10 . = = N2 162 2 N 2 16
But
T λ2 12 3 = 1 = = 16 λ1 T2 4
or
λ2 =
3 λ1 4
1 3 1 A2 = λ1 × N10 × 8 4 3
=
1 1 A10 λ1 N10 = 32 32
= (1/32) µCi
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