MCQ
Mercury violet light $(\lambda = 4558{Å})$ is falling on a photosensitive material $(\phi = 2.5eV)$. 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.
 $2.65 \times {10^5}$
b
(b) By using$E = {W_0} + \frac{1}{2}mv_{\max }^2$; where $E = \frac{{12375}}{{4558}} = 2.71\;eV$

$ \Rightarrow 2.71\;eV = 2.5\;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}\;m/s$

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