In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is :

1.8 V

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1.3 V

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0.5 V

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2.3 V

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Solution

The correct option is C 0.5 V
Let $V_{0}$ be the stopping potential

In a photoelectric effect, the maximum kinetic energy of electron ,

$K_{m a x} = e V_{0}$

so, $V_{0} = \frac{K_{m a x}}{e} = \frac{0.5 e V}{e} = 0.5 V$ (as eV=1 electronic charge times 1 Volt)

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In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is :

The correct option is C 0.5 VLet V0 be the stopping potentialIn a photoelectric effect, the maximum kinetic energy of electron , Kmax=eV0so, V0=Kmaxe=0.5eVe=0.5V (as eV=1 electronic charge times 1 Volt)

The correct option is C 0.5 V
Let $V_{0}$ be the stopping potential

In a photoelectric effect, the maximum kinetic energy of electron ,

$K_{m a x} = e V_{0}$

so, $V_{0} = \frac{K_{m a x}}{e} = \frac{0.5 e V}{e} = 0.5 V$ (as eV=1 electronic charge times 1 Volt)