MCQ
Mercury violet light $(\lambda=4558\  \mathring A)$ is falling on a photosensitive material $(\phi=2.5 \mathrm{eV})$. The speed of the ejected electrons is in $m s^{-1}$, about
  • A
    $3 \times 10^5$
  • $2.65 \times 10^5$
  • C
    $4 \times 10^4$
  • D
    $3.65 \times 10^7$

Answer

Correct option: B.
$2.65 \times 10^5$
$ \text { By using } E=W_0+\frac{1}{2} m v_{\max }^2 ; \text { where } E=\frac{12375}{4558}=2.71 \mathrm{eV} $
$ \Rightarrow 2.71 \mathrm{eV}=2.5 \mathrm{eV}+\frac{1}{2} \times 9.1 \times 10^{-31} \times v_{\max }^2 $
$ \Rightarrow 0.21 \times 1.6 \times 10^{-19}=\frac{1}{2} \times 9.1 \times 10^{-31} \times v_{\max }^2 $
$ \Rightarrow v_{\max }=2.65 \times 10^5 \mathrm{~m} / \mathrm{s}$

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