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An electron in a hydrogen atom makes a transition from second excited state to ground state. The emitted photon is made to incident on a photo metal of work function 3.1 eV. The minimum possible de-Broglie’s wavelength of emitted electron from the photo metal is

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$10.336$ Å

$13.78$ Å

$15$ Å

$4.08$ Å

answer is D.

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Detailed Solution

$E = 13.6 - \frac{13.6}{9} = 12.09 e V$

$K E = h v - w_{o} = 12.09 - 3.1 = 8.99 e V$

$λ_{e} = \frac{12.27}{\sqrt{8.99}} = 4.08 A °$

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