The light rays having photons of energy $4.2 \mathrm{eV}$ are falling on a metal surface having a work…

The light rays having photons of energy $4.2 \mathrm{eV}$ are falling on a metal surface having a work function of $2.2 \mathrm{eV}$. The stopping potential of the surface is:
  1. $2 \mathrm{eV}$
  2. $2 \mathrm{~V}$
  3. $1.1 \mathrm{~V}$
  4. $6.4 \mathrm{~V}$

Solution

By Einstein photoelectric equation, we have; $\begin{aligned} \text { K.E. } & =h v-\phi \\ e \mathrm{~V}_0 & =4.2 \mathrm{eV}-2.2 \mathrm{eV} \\ e \mathrm{~V}_0 & =2 \mathrm{eV}=2 \text { volt } \end{aligned}$

Asked in: NEET 2022 (Phase 2)

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