Question

Light from a discharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photoelectrons emitted from sodium is $0.73eV$, The work function for sodium is $1.82eV$. Find the recoil speed of the emitting atom assuming it to be at rest before the transition.
(Ionization potential of hydrogen is $13.6eV$)

• A. $0.9m/\sec$.
• B. $0.8m/\sec$.
• C. $1m/\sec$.
• D. $2m/\sec$.

Answer 1

Answer: (B)

$0.8m/\sec$.

$\text{Solution}:$

We know that

$E_0=\frac{-13.6}{n^2}\frac{eV}{atom}$ for atom

Let electron jump from $n_2$ to $n_1$

then

$E_{n_2}-E_{n_1}=\frac{13.6}{n_2^2}-\frac{13.6}{n_2^2}$

$⟹2.55=13.6\frac{1}{n_1^2}-\frac{1}{n_2^2}$

By hit and trial we get $n_2=4$ and $n_1=2$ [angular momentum m v r = $\frac{nh}{2\pi }$]

Change in angular momentum = $\frac{m_{1}h}{2\pi }-\frac{m_{2}h}{2\pi }=\frac{h}{2\pi }(2-4)=\frac{h}{2\pi }\times (-2)=\frac{h}{\pi }$

The momentum of emitted photon can be found by de Broge relationship

$\lambda \frac{h}{p}$ $⟹P=\frac{h}{\lambda }=\frac{hv}{c}=\frac{E}{c}$

$\therefore \frac{2.55\times 1.6\times {10}^{-19}}{3\times {10}^8}=0.8m/sec$

$\text{Hence B is the correct option}$