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A neutron of kinetic energy 65eV collides inelastically with a singly ionized helium atom at rest. It is scattered at an angle of 90° with respect of its original direction

(i) Find the allowed valued of the energy of the neutron and that of the atom after the collision.
(ii) If the atom get de-excited subsequently by emitting radiation, find the frequencies of the emitted radiation.
[Given : Mass of He atom = 4 x (mass of neutron), Ionization energy of H atom = 13.6 eV]

Radhika Batra , 11 Years ago
Grade 11
anser 1 Answers
Kevin Nash
Hello Student,
Please find the answer to your question
Applying conservation of linear momentum in horizontal direction
(Initial Momentum )x = (Final Momentum)x
(P1)x = (Pf)x
⇒ √2Km = √2(4m)K1 cos θ …(i)
Now applying conservation of linear momentum in Y – direction
(Pi)y = (Pf)y
  1. = √2K2 m - √ 2 (4m) K1 sin θ
⇒ √ 2 K2 m = √2(4m)K1 sin θ …. (ii)
Squaring and adding (i) and (ii)
2Km + 2Km2 m = 2 (4m) K1 + 2 (4m)K1
K1 + K2 = 4K1 ⇒ K = 4 K1 – K2 ⇒ 4K1 – K2 = 65 …(iii)
When collision takes place, the electron gains energy and jumps to higher orbit.
Applying energy conservation
K = K1 + K2 + ∆E
⇒ 65 = K1 + K2 + ∆E ….(iv)
Possible value of ∆E for He+
Case (1)
∆E1 = - 13. 6 – (- 54.4 ) = 40.8 eV
⇒ K1 + K2 = 24.2 eV from (4)
Solving with (3), we get
K2 = 6.36 eV ; K1 = 17 .84 eV
Case (2)
∆E2 = - 6.04 – (- 54 .4) = 48 .36 eV
⇒ K1 + K2 = 16.64 eV from (4)
Solving with (3), we get K2 = 0.312 eV; K1 = 16. 328 eV
Case (3)
∆E3 = - 3.4 – (- 54.4) = 51.1 eV
⇒ K1 + K2 = 14 eV
Solving with (3), we get
K2 = 15.8 eV ; K1 = - 1.8 eV
But K.E. can never be negative therefore case (3) is not possible.
Therefore, the allowed values of kinetic energies are only that of case (1) and case (2) and electron can jump upto n = 3 only.
(ii) Thus when electron jumps back there are three possibilities
n3 ⟶ n1 or n3 ⟶ n2 and n2 ⟶ n1
The frequencies will be
y1 = E3 – E2 /h ; v2 = E3 – E1 / h ; v3 = E2 – E1 / h
i.e., 1.82 x 1015 Hz; 11.67 x 1015 Hz; 9.84 x 1015 Hz
Thanks
Kevin Nash
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Last Activity: 11 Years ago
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