Question

The maximum kinetic energy of an electron emitted due to photoelectric effect is $20\text{nJ}$. For the same incident radiation and same metal surface at same physical conditions, the stopping potential of the electron would be,

• A. $1.25\times {10}^9\text{V}$
• B. $125\text{MV}$
• C. $125\text{GV}$
• D. $125\text{KV}$

Answer 1

The correct option is C $125\text{GV}$

As we know,
${(K.E)}_{max}=E-{\varphi }_0$
And, $e{V}_s=E-\varphi {}_0$, where $V_s$ is the stopping potential.

$\therefore {(K.E)}_{max}=e{V}_S$

$\Rightarrow V_S=\frac{K.E_{max}}{e}=\frac{20\times {10}^{-9}}{1.6\times {10}^{-19}}$

$\Rightarrow V_S=1.25\times {10}^{11}\text{V}=125\text{GV}$

Hence, $\left(C\right)$ is the correct answer.

Why this question? This is a typical NEET question where the following concepts are tested $K.E_{max}=e{V}_S$ where, $V_S\Rightarrow$ stopping potential.