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Photons and Photoelectric Effect

Question

The photoelectric threshold for a metal surface is $$6600\ \overset {o}{A}$$. The work function for this is

A

$$1.87\ V$$

B

$$1.87\ eV$$

C

$$18.7\ eV$$

D

$$0.18\ eV$$

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Solution

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Correct option is B. $$1.87\ eV$$
As we know,
$$W_0=\dfrac {12375}{6600}=1.87\ eV$$

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