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.
If the atom gets de-excited subsequently by emitting radiation, find the frequencies of the emitted radiation.
[Given : Mass of $Heatom=4\times$ (mass of neutron) Ionization energy of $H$ atom $=13.6eV]$.

• A. $82\times {10}^{15}Hz,11.67\times {10}^{15}Hz,9.84\times {10}^{15}Hz$.
• B. $8\times {10}^{15}Hz,11.67\times {10}^{15}Hz,9.84\times {10}^{15}Hz$.
• C. $1.82\times {10}^{15}Hz,11.67\times {10}^{15}Hz,12\times {10}^{15}Hz$.
• D. $1.82\times {10}^{15}Hz,11.67\times {10}^{15}Hz,9.84\times {10}^{15}Hz$.

Answer 1

The correct option is D $1.82\times {10}^{15}Hz,11.67\times {10}^{15}Hz,9.84\times {10}^{15}Hz$.

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

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

$=\sqrt{2}K2m-\sqrt{2}\left(4m\right)K1sin\theta$

$\to \sqrt{2}K2m=\sqrt{2}\left(4m\right)K1sin\theta \dots .\left(ii\right)$

Squaring and adding (i) and (ii)

$2Km+2Km2m=2\left(4m\right)K1+2\left(4m\right)K1$

$K1+K2=4K1\Rightarrow K=4K1–K2\Rightarrow 4K1–K2=65\dots \left(iii\right)$

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

Applying energy conservation

$K=K1+K2+\Delta E$

$65=K1+K2+\Delta E$

Possible value of $\Delta E$ for He+

Case (1)

$\Delta E1=-13.6–(-54.4)=40.8eV$

$K1+K2=24.2eV$ from (4)

Solving with (3), we get

$K2=6.36eV;K1=17.84eV$

Case (2)

$\Delta E2=-6.04–(-54.4)=48.36eV$

$K1+K2=16.64eV$ from (4)

Solving with (3), we get $K2=0.312eV;K1=16.328eV$

$\Delta E3=-3.4–(-54.4)=51.1eV$

$\Rightarrow K_1+K_2=14eV$

Solving with (3), we get

$K_2=15.8eV;K_1=-1.8eV$

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 $\to$ n1 or n3 $\to$n2 and n2 $\to$ n1

The frequencies will be

$y1=E3–E2/h$

$v2=E3–E1/h$

$v3=E2–E1/h$

$1.82$ x ${10}^{15}Hz;$

$11.67$ x ${10}^{15}Hz;$

$9.84$ x ${10}^{15}Hz$

Hence the correct option:D