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Wave Optics

Question

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The maximum kinetic energy of a photoelectron is 3 eV. What is its stopping potential?

Solution

Maximum K.E of particle = 3 eV

Since, $K . E ._{\max} = {eV}_{0}$

$\Rightarrow$ $3 \times 1.6 \times 10^{- 19} = 1.6 \times 10^{- 19} V_{0}$

$∴$ Stopping potential, $V_{0} = 3 V .$

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