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Standard XII

Physics

Stopping Potential

The maximum k...

Question

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy $6$ eV fall on it is $4$ eV. The stopping potential is?

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Solution

The correct option is D $4$ V
$V_{0} = \frac{K_{m a x}}{e}$
$V_{0} = \frac{4_{e v}}{e} = 4 v$

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The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential is?

The correct option is D 4VV0=KmaxeV0=4eve=4v

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy $6$ eV fall on it is $4$ eV. The stopping potential is?

The correct option is D $4$ V
$V_{0} = \frac{K_{m a x}}{e}$
$V_{0} = \frac{4_{e v}}{e} = 4 v$