Light of energy 2 . 0 eV falls on a metal of work function 1 . 4 eV . The stopping potential is

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TestHub · Accepted Answer

From Planck's quantum theory, the maximum kinetic energy ( $E_{k}$ ) of photoelectron emitted from the metal is

$E_{k} = h ν - W$

Where $W$ is the work function of metal and $h ν$ is the energy of the photon absorbed by the metal.

Given, $h ν = 2 eV, W = 1.4 e V$

$∴ E_{k} = 2 - 1.4 = 0.6 eV$

Hence, stopping potential is , $V_{s} = \frac{E}{e} = \frac{0.6 e V}{e} = 0.6 V$

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