Question

The stopping potential for the photoelectrons emitted from a metal surface of work function $1.7\text{eV}$ is $10.4\text{V}.$ Find the wavelength of the radiation used. Also identify the energy levels in hydrogen atom which will emit this wave length.

• A. $2044\mathring{A};n=3\text{to}n=1$
• B. $2044\mathring{A};n=2\text{to}n=1$
• C. $1024\mathring{A};n=2\text{to}n=1$
• D. $1024\mathring{A};n=3\text{to}n=1$

Answer 1

The correct option is D $1024\mathring{A};n=3\text{to}n=1$

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

From, Einstein's photo electric equation,

$E_{in}={\varphi }_0+e{V}_0$

$=1.7+10.4=12.1\text{eV}$

$E_{in}=\frac{hc}{\lambda }\Rightarrow \lambda =\frac{12400}{E_{in}}\mathring{A}$

$\Rightarrow \lambda =\frac{12400}{12.1}\approx 1024\mathring{A}$

As we know, H-atom discrete energy levels as,

$E_1=-13.6\text{eV};E_3=-1.5\text{eV}$

$\Rightarrow E_3-E_1=-1.5+13.6=12.1\text{eV}$

Hence, the energy level is corresponding to,

$\therefore \text{From}n=3\text{to}n=1$

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