Light of wavelength $4000 \circ A$ falls on a metal surface having work function of 2.0 eV. The stopping potential is

3.1 V

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2.0 V

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5.1 V

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1.1 V

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Solution

The correct option is D
1.1 V

Energy of incident photon

$= \frac{h c}{λ} = \frac{12400}{λ} e V = \frac{12400}{4000} e V = 3.1 e V$

$k_{max} = h v - ϕ = 3.1 e V - 2.0 e V = 1.1 e V$ . Hence, stopping potential 1.1V

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Light of wavelength 4000 ∘A falls on a metal surface having work function of 2.0 eV. The stopping potential is

The correct option is D 1.1 V Energy of incident photon =hcλ=12400λeV=124004000eV=3.1 eV kmax=hv−ϕ=3.1eV−2.0eV=1.1eV. Hence, stopping potential 1.1V

Light of wavelength $4000 \circ A$ falls on a metal surface having work function of 2.0 eV. The stopping potential is

The correct option is D
1.1 V

Energy of incident photon

$= \frac{h c}{λ} = \frac{12400}{λ} e V = \frac{12400}{4000} e V = 3.1 e V$

$k_{max} = h v - ϕ = 3.1 e V - 2.0 e V = 1.1 e V$ . Hence, stopping potential 1.1V