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Question: The stopping potential for a photelectric emission process is 10 V. The maximum kinetic energy of th...

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 e = 1.6×10−191.6 \times 10^{-19}C ]

Answer

1.6×10−18 J1.6 \times 10^{-18} \, \text{J}

Explanation

Solution

The maximum kinetic energy KmaxK_{\text{max}} of the emitted electrons is given by:

Kmax=eVstopK_{\text{max}} = eV_{\text{stop}}

Substitute the given values:

Kmax=(1.6×10−19 C)×(10 V)=1.6×10−18 JK_{\text{max}} = (1.6 \times 10^{-19} \, \text{C}) \times (10 \, \text{V}) = 1.6 \times 10^{-18} \, \text{J}