A photon of energy 4 eV is incident on a metal surface whose work function is 2 eV. Calculate the minimum reverse potential to be applied for stopping the emission of electrons.

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Solution

$(\frac{m v^{2}}{2})_{m a x} = h ν - ϕ$

$= 4 e V - 2 e V = 2 e V$

Now, $e V_{0} = 2 e V$

$V_{0} = 2 V$

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A photon of energy 4 eV is incident on a metal surface whose work function is 2 eV. Calculate the minimum reverse potential to be applied for stopping the emission of electrons.

(mv22)max=hν−ϕ=4eV−2eV=2eVNow, eV0=2eVV0=2V

$(\frac{m v^{2}}{2})_{m a x} = h ν - ϕ$

$= 4 e V - 2 e V = 2 e V$

Now, $e V_{0} = 2 e V$

$V_{0} = 2 V$