Atomic StructureHard

Question

Photoelectrons are liberated by ultraviolet light of wavelength 3000 Å from a metallic surface for which the photoelectric threshold is 4000 Å. The de Broglie wavelength of electrons emitted with maximum kinetic energy is

Options

A.1000 Å
B.42.43 Å
C.12.05 Å
D.3.54 Å

Solution

K.E. of electrons = $\frac{12400}{3000} - \frac{12400}{4000} = 1.03\text{ eV}$

$\therefore\lambda = \sqrt{\frac{150}{1.03}} = 12.05\overset{o}{A}$

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