Q: An electron with initial kinetic energy of 100 eV is acceleration through a potential difference of 50 V. Now the de-Broglie wavelength of electron becomes.

(A) 1 A°

(B) $\sqrt{1.5}$ A°

(C) $\sqrt{3}$ A°

(D) 12.27 A°

Solution : Kinetic Energy , K = 100 eV + 50 eV = 150 e V

$\large \lambda = \sqrt{\frac{150}{V}}$

$\large \lambda = \sqrt{\frac{150}{150}}$

= 1 A°

Correct option is (A)