Question

The stopping potential for the photoelectrons emitted from a metal surface of work function 1.7 eV is 10.4 V. Find the wavelength of the radiation in $\stackrel{0}{A}$

Answer 1

From the Einstein's photo electric equation, we get

$\frac{hc}{\lambda }=e{V}_0+\varphi$

$\therefore \frac{hc}{\lambda }=10.4eV+1.7eV$

$\therefore \frac{hc}{\lambda }=12.1eV$

$\therefore \lambda =\frac{hc}{12.1\times 1.6\times {10}^{-19}}$

$=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{12.1\times 1.6\times {10}^{-19}}$

$=1026A°$

Now, the transition between $3\to 1$

we get

$\frac{1}{{\lambda }_1}=R(\frac{1}{1^2}-\frac{1}{3^2})=R(1-\frac{1}{9})$

$=\frac{8}{9}R=\frac{8\times 1.097\times {10}^7}{9}$

$\therefore \lambda =1026A°$

$\therefore$The energy level in hydrogen atom which will emit this wavelength is $n=3$ to $n=1$