Question

A neutron of kinetic energy $65eV$ collides inelastically with a singly ionized helium atom at rest. It is scattered at an angle of ${90}^{\circ }$ with respect to its original direction.
Find the allowed values of the energy of the neutron and that of the atom after the collision.
[Given : Mass of $Heatom=4\times$ (mass of neutron) Ionization energy of $H$ atom $=13.6eV$]

• A. $7.36eV,0.312,17.8eV;16.328eV$.
• B. $6.36eV,0.312,17.8eV;16.328eV$.
• C. $6.36eV,0.312,87.8eV;16.328eV$.
• D. $6.36eV,0.312,17.8eV;26.328eV$.

Answer 1

Answer: (A)

$7.36eV,0.312,17.8eV;16.328eV$.

solution:

Applying conservation of linear momentum in horizontal direction

(Initial Momentum )x = (Final Momentum)x

$(P1{)}_x=(Pf{)}_x$

$\Rightarrow \sqrt{2}Km=\sqrt{2}\left(4m\right)K_{1}cos\theta \dots \left(i\right)$

Now applying conservation of linear momentum in Y – direction

$(Pi{)}_y=(Pf{)}_y$

$=\sqrt{2}{K}_{2}m-\sqrt{2}\left(4m\right)K_{1}sin\theta$

$\to \sqrt{2}{K}_{2}m=\sqrt{2}\left(4m\right)K_{1}sin\theta \dots .\left(ii\right)$

Squaring and adding (i) and (ii)

$2Km+2Km2m=2\left(4m\right)K_1+2\left(4m\right)K_1$

$K_1+K_2=4K_1\Rightarrow K=4K_1–K_2\Rightarrow 4K_1–K_2=65\dots \left(iii\right)$

When collision takes place, the electron gains energy and jumps to higher orbit.

Applying energy conservation

$K=K_1+K_2+\Delta E$

$65=K_1+K_2+\Delta E$

Possible value of $\Delta E$ for He+

Case (1)

$\Delta E_1=-13.6–(-54.4)=40.8eV$

$K_1+K_2=24.2eV$ from (4)

Solving with (3), we get

$K_2=6.36eV;K_1=17.84eV$

Case (2)

$\Delta E_2=-6.04–(-54.4)=48.36eV$

$K_1+K_2=16.64eV$ from (4)

Solving with (3), we get $K_2=0.312eV;K_1=16.328eV$

Hence the correct option: A