Q155 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Physics

Let $v_1$ and $v_2$ be the maximum velocities of the emitted electrons when the surface of a metal is illuminated with light waves of energy $E_1=4 \mathrm{eV}$ and $E_2=2.5 \mathrm{eV}$, respectively. If the work function of the metal is $2 \mathrm{eV}$, then the ratio $\frac{v_1}{v_2}$ is

  • A
    1.6
  • B
    4
  • Correct answer
    C
    2
  • D
    0.5

Solution

In case of photoemission, $K_{\max }=\frac{1}{2} m v_{\max }^2=h f-\phi_0$ For photons of energies $4 \mathrm{eV}$ and $2.5 \mathrm{eV}$, we have ( given, work function $=2 \mathrm{eV}$ ) $\frac{1}{2} m v_1^2=4-2=2$ $\ldots(i)$ and $\quad \frac{1}{2} m v_2^2=2.5-2=0.5$ $\ldots(ii)$ Dividing Eq. (i) by Eq. (ii), we get $\frac{v_1^2}{v_2^2}=\frac{2}{0.5}=4 \Rightarrow \frac{v_1}{v_2}=2$

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

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