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Light of wavelength $4000 A^{\circ}$ is incident on a potassium surface whose work function is $2.1 ev$ . What will be the stopping potential

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1 v

2.1 v

4000 v

3.1 v

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detailed solution

Correct option is A

Wavelength λ=4000 A∘ Work function w=2.1 ev E=w+ev0 12,4004000=2.1+ev0 (∵E=12400λ A∘) 3.1=2.1+ev0 v0=1 v

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Light of wavelength 4000 A∘ is incident on a potassium surface whose work function is 2.1 ev . What will be the stopping potential

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Light of wavelength $4000 A^{\circ}$ is incident on a potassium surface whose work function is $2.1 ev$ . What will be the stopping potential