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 = $\frac{12400}{3000} - \frac{12400}{4000} = 1.03\text{ eV}$

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

Create a free account to view solution

View Solution Free
Topic: Atomic Structure·Practice all Atomic Structure questions

More Atomic Structure Questions

The wavelength of a spectral line obtained by an electronic transition is inversely proportional to...3.011 × 1022 atoms of an element weight 1.15gm. The atomic mass of the element is :-...In two hydrogen atoms A and B, the electrons are revolving around the nucleus in circular orbits of radius r and 4r, res...The uncertainties in the velocities of two particle, A and B are 0.05 and 0.02 ms-1 respectively. The mass of B is five ...The number of nodal planes in a Px orbotal is :...