Question

A free hydrogen atom in ground state is at rest. A neutron of kinetic energy K collides with the hydrogen atom. After collision hydrogen atom emits two photons in succession one of which has energy $2.55eV$. Assume that the hydrogen atom and neutron has same mas.

• A. Minimum value of K is $25.5eV$
• B. Minimum value of K is $12.75eV$
• C. The other photon has energy $10.2eV$
• D. Th upper energy level is of excitation energy $12.5eV$

Answer 1

Answer: (A), (C)

Minimum value of K is $25.5eV$
The other photon has energy $10.2eV$

$\text{Let collision be inelastic}$

$\begin{array}{c}\text{by conservation of Linear momentum}\\ \begin{array}{cc}mv+0& =2m{v}^{\prime }\\ v^{\prime }& =\frac{v}{2}\end{array}\end{array}$

$\begin{array}{c}\text{Now,}\\ \text{initial kinetic Energy given}=K=\frac{1}{2}m{v}^2\\ \text{and final}KE=\frac{1}{2}\left(2m\right)v^{\prime 2}\end{array}$

$\begin{array}{cc}& =\frac{1}{2}2m\times \frac{v^2}{4}\\ {\left(K_E\right)}_{\text{tinal}}& =\frac{k}{2}\end{array}$

$\begin{array}{cc}\text{Change in kinetic Energy}& =k-\frac{k}{2}\\ & =\frac{k}{2}\end{array}$

$\begin{array}{c}\text{This change in kinetic energy leads to}\\ \text{Excitation of two photon of which one}\\ \text{have}2.55\text{ev energy.}\end{array}$

$\begin{array}{c}\text{from Energy level of hydrogen}\\ \begin{array}{cc}\Delta E=& E_4-E_2\\ & =-0.85-(-3.41)\\ & =2.55ev\end{array}\end{array}$

$\begin{array}{c}\text{So, we can say when collision take place}\\ \text{the electron from ground state excite to}{4}^{\text{th}}\\ \text{energy level and after de-excitation one}\\ \text{electron is de-excite from}4\text{to}2\text{, then}\\ \text{other photon should from}2\text{to}1\end{array}$

$\begin{array}{c}{E}_2-E_1\\ =-3\cdot 4-(-13\cdot 6)\\ E_{p2}=10.2ev\end{array}$

$\begin{array}{cc}\text{Now,}& \\ \frac{K}{2}& =E{p}_1+E_2\\ & =2.55+10.2ev\\ \frac{K}{2}& =12.75ev\\ K& =25.5ev.\end{array}$

$\text{option (c) is also correct}$