A light having wavelength 300 nm falls on a metal surface. Work function of metal is 2.54 eV. What is stopping potential?

2.3 V

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2.59 V

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1.60 V

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1.20 V

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Solution

The correct option is C
1.60 V

$e V_{0} = \frac{h c}{λ} - W_{0}$

$= \frac{1240 e V}{300} - 2.54 e V$

$= 4.13 - 2.54 = 1.59 e V$

$V_{0} = 1.6 V$

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Similar questions

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

2000 ˙ A

4.2 eV

(h c = 12400 eV ˙ A)

100 A^{0}

500 A^{0}

0.40 μ A

λ

\frac{h c}{λ_{0}}

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