Question

The kinetic energy of a proton is increased by nine times, the wavelength of the de-Broglie wave associated with it would become:

• A. 3 times
• B. 9 times
• C. $\frac{1}{3}$ times
• D. $\frac{1}{9}$ times

Answer 1

The correct option is C $\frac{1}{3}$ times

The relation between de-Broglie wavelength and kinetic energy is given by,
$\lambda =\frac{h}{\sqrt{2m(K.E.)}}$
$K.E{.}^{\prime }$= $9K.E.$
Then,
${\lambda }^{\prime }=\frac{h}{\sqrt{2m(K.E^{\prime }.)}}$
$=\frac{h}{\sqrt{2m(9K.E.)}}$
$=\frac{h}{3\sqrt{2m(K.E.)}}$
$=\frac{\lambda }{3}$