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MHT CET · Physics · Dual Nature of Matter

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. A \(3.2 \times 10^{-19} \mathrm{~J}\)
  2. B \(1.6 \times 10^{-19} \mathrm{~J}\)
  3. C \(1.6 \times 10^{-18} \mathrm{~J}\)
  4. D 0 J
Verified Solution

Answer & Solution

Correct Answer

(C) \(1.6 \times 10^{-18} \mathrm{~J}\)

Step-by-step Solution

Detailed explanation

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