Q151 / 180 · KCET 2013 · Physics

Maximum velocity of the photoelectron emitted by a metal is $1.8 \times 10^{6} \mathrm{~ms}^{-1}$. 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

Given, $v=1.8 \times 10^{6} \mathrm{~m} / \mathrm{s} ; \frac{e}{m}=1.8 \times 10^{11} \mathrm{C} / \mathrm{kg}$ We have, $\quad e V_{0}=\frac{1}{2} m v^{2}$ $\begin{array}{ll} \Rightarrow & V_{0} \frac{e}{m}=\frac{v^{2}}{2} \\ \Rightarrow \quad & V_{0} \times 1.8 \times 10^{11}=\frac{1.8 \times 1.8 \times\left(10^{6}\right)^{2}}{2} \\ \Rightarrow \quad & V_{0}=9 \mathrm{~V} \end{array}$

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

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