The stopping potential for a photelectric emission process is 10 V . The maximum kinetic energy of the…

The stopping potential for a photelectric emission process is 10 V . The maximum kinetic energy of the electrons ejected in the process is [Charge on electron $\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ ]
  1. $3.2 \times 10^{-19} \mathrm{~J}$
  2. $1.6 \times 10^{-19} \mathrm{~J}$
  3. $1.6 \times 10^{-18} \mathrm{~J}$
  4. 0 J

Solution

Maximum kinetic energy is given by, $\begin{aligned} & \text { (K.E. })_{\text {max }}=\mathrm{eV}_{\mathrm{s}} \\ & \text { (K.E. })_{\text {max }}=\left(1.6 \times 10^{-19}\right) \times 10 \\ & \text {...(given, } \mathrm{V}_{\mathrm{s}}=10 \mathrm{~V} \text { ) } \\ & \therefore \quad(\text { K.E. })_{\max }=1.6 \times 10^{-18} \mathrm{~J} \end{aligned}$

Asked in: MHT CET 2024 (02 May Shift 2)

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