Q59 / 75 · VITEEE 2021 · Physics

Photoelectric work function of a metal is $1 \mathrm{eV}$. Light of wavelength $\lambda=3000 Å$ falls on it. The photo electrons come out with a maximum velocity of :

  • A
    10 metres/sec
  • B
    $10^2$ metres $/ \mathrm{sec}$
  • C
    $10^4$ metres $/ \mathrm{sec}$
  • Correct answer
    D
    $10^6$ metres $/ \mathrm{sec}$

Solution

$$ \begin{aligned} & \text { } h v=W+\frac{1}{2} m v^2 \quad \text { or } \frac{h c}{\lambda}=W+\frac{1}{2} m v^2 \\ & \text { Here } \lambda=3000 Å=3000 \times 10^{-10} \mathrm{~m} \\ & \text { and } \mathrm{W}=1 \mathrm{eV}=1.6 \times 10^{-19} \text { joule } \end{aligned} $$ $$ \begin{aligned} & \therefore \frac{\left(6.6 \times 10^{-34}\right)\left(3 \times 10^8\right)}{3000 \times 10^{-10}} \\ & =\left(1.6 \times 10^{-19}\right)+\frac{1}{2} \times\left(9.1 \times 10^{-31}\right) \mathrm{v}^2 \end{aligned} $$ Solving we get, $v \cong 10^6 \mathrm{~m} / \mathrm{s}$

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

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