Two radiations of photon energies $1 \mathrm{eV}$ and $2.5 \mathrm{eV}$, successively illuminate a…

Two radiations of photon energies $1 \mathrm{eV}$ and $2.5 \mathrm{eV}$, successively illuminate a photosensitive metallic surface of work function $0.5 \mathrm{eV}$. The ratio of the maximum speeds of the emitted electrons is
  1. $1: 4$
  2. $1: 2$
  3. $1: 1$
  4. $1: 5$

Solution

We have $\frac{1}{2} m v_{\max }^2=E-\phi$ Here case (i) $\begin{gathered} \frac{1}{2} m v_{1 \max }^2=(1-0.5) \mathrm{eV} \\ \frac{1}{2} m v_{2 \max }^2=(2.5-0.5) \mathrm{eV} \end{gathered}$ Hence, $\quad \frac{v_{1 \max }^2}{v_{2 \max }^2}=\frac{1}{4}$ $\frac{v_{1 \max }}{v_{2 \max }}=\frac{1}{2}$

Asked in: NEET 2012 (Mains)

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