Q107 / 180 · AIIMS 2018 (27 May) · Physics

Maximum velocity of the photoelectron emitted by a metal is $1.8 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}$. Take the value of specific charge of the electron is $1.8 \times 10^{11} \mathrm{C} \mathrm{kg}^{-1}$. Then the stopping potential in volt is

  • A
    $1$
  • B
    $3$
  • Correct answer
    C
    $9$
  • D
    $6$

Solution

In photoelectic effect stopping potenital is given by $V_S=\frac{1}{2} \frac{m}{e} v_{\max }^2$ Here, $\begin{aligned} & \frac{e}{m}=1.8 \times 10^{11} \mathrm{C} \mathrm{kg}^{-1}, v_{\max }=1.8 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1} \\ & \therefore \quad V_S=\frac{1}{2} \times \frac{1}{1.8 \times 10^{11}} \times\left(1.8 \times 10^6\right)^2=9 \mathrm{~V}\end{aligned}$

Hence, the correct answer is option: 'C'.

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