Question

The stopping potential for the photo electrons emitted from a metal surface of work function $1.7\text{eV}$ is $10.4\text{V}.$ Identify the energy levels corresponding to the transitions in hydrogen atom which will result in emission of wavelength equal to that of incident radiation for the above photoelectric effect.
$[$Take $hc=1240\text{eV. nm}]$

• A. $n=3$ to $n=1$
• B. $n=3$ to $n=2$
• C. $n=2$ to $n=1$
• D. $n=4$ to $n=1$

Answer 1

The correct option is A $n=3$ to $n=1$

Given,
$V_0=10.4\text{V},{\varphi }_0=1.7\text{eV}$

If incident radiation has wavelength $\lambda$, then

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

$\Rightarrow \frac{hc}{\lambda }=10.4+1.7=12.1\text{eV}$

$\lambda =\frac{1240\text{eV (nm)}}{12.1\text{eV}}.....\left(1\right)$

And for Hydrogen atom, energy levels are,

$E_n=-13.6{\left(\frac{Z}{n}\right)}^2$

$\Delta E=-(E_f-E_i)=E_i-E_f$

For transition $n=3\to n=1$

$\Delta E=E_3-E_1=-1.51-\left(13.6\right)\approx 12.1\text{eV}$

$\because \Delta E=\frac{hc}{\lambda }$

$\Rightarrow \lambda =\frac{1240\text{eV (nm)}}{12.1\text{eV}}.....\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, energy of transition form $n=3$ to $n=1$ corresponds to energy of incident radiation.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.