Question

The stopping potential for photo-electrons ejected from a surface is $4.5\text{V}$ when $155\text{nm}$ photon is incident on the surface. Calculate the stopping potential when $248\text{nm}$ photon is incident on the surface. $(\text{Use}hc=1240\text{eV nm})$

• A. $1.5\text{V}$
• B. $2.5\text{V}$
• C. $2.0\text{V}$
• D. $1.2\text{V}$

Answer 1

The correct option is A $1.5\text{V}$

Given: ${\lambda }_1=155\text{nm}$ ${\lambda }_2=248\text{nm}$
$V_1=4.5\text{eV}$ $V_2=?$

The Formula for stooping potential is given by,

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

$\Rightarrow e{V}_1=\frac{hc}{{\lambda }_1}-\varphi$

$\Rightarrow 4.5\text{eV}=\frac{1240\text{eV nm}}{155\text{nm}}-\varphi$

$\Rightarrow \varphi =3.5\text{eV}$

$\therefore e{V}_2=\frac{hc}{{\lambda }_2}-\varphi$

$\Rightarrow e{V}_2=\frac{1240\text{eV nm}}{248\text{nm}}-3.5\text{eV}$

$\Rightarrow e{V}_2=5\text{eV}-3.5\text{eV}=1.5\text{eV}$

$\Rightarrow V_2=1.5\text{V}$

Hence, option $\left(A\right)$ is correct.