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Q.

The light rays having photons of energy $10.2 e V$ are falling on a metal surface having a work function $1.2 e V$ . The minimum de-Broglie wavelength of electrons emitted from the metal will be

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a

$4 A^{0}$

b

$400 A^{0}$

c

$40 A^{0}$

d

$0.4 A^{0}$

answer is A.

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Detailed Solution

$K E_{\max} = (10.2 + 1.2) e V = 9 e V$
$λ = \frac{12.27}{\sqrt{9}} A^{0} = 4.09 A^{0}$

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