The photo-electric threshold wavelength for a metal is 6620 $A^{o}$ . Its work function is (given $h = 6.62 \times 10^{- 34} J s, c = 3 \times 10^{8}$ m/s)

1.87 eV

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2.00 eV

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3.00 eV

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2.87 eV

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Solution

The correct option is A 1.87 eV
$ϕ = \frac{h c}{λ_{T}}$

$= \frac{1240}{662} (h c = 1240 e V - n m)$

$= 1.87 e V$
So, the answer is option (A).

Suggest Corrections

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(h = 6.6 \times 10^{- 34} J s, c = 3 \times 10^{8} m s^{- 1}, 1 e V = 1.6 \times 10^{- 19} J)

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