What is the stopping potential when the metal with work function 0.6 eV is illuminated with the light of 2 eV

2.6 V

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3.6 V

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0.8 V

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1.4 V

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Solution

The correct option is D
1.4 V

$V_{0} = \frac{(E - W_{0})}{e} = \frac{(2 e V - 0.6 e V)}{e} = 1.4 V$

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What is the stopping potential when the metal with work function 0.6 eV is illuminated with the light of 2 eV

The correct option is D 1.4 V V0=(E−W0)e=(2eV−0.6eV)e=1.4V

The correct option is D
1.4 V

$V_{0} = \frac{(E - W_{0})}{e} = \frac{(2 e V - 0.6 e V)}{e} = 1.4 V$