Question

If the kinetic energy of the particle is increased by 16 times, the percentage change in the de Broglie wavelength of the particle is :

• A. 25%
• B. 75%
• C. 60%
• D. 50%

Answer 1

Answer: (B)

75%

Answer 2

: As $\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mK}}or\lambda \propto \frac{1}{\sqrt{K}}$

$\therefore\frac{\lambda'}{\lambda} = \sqrt{\frac{K}{K'}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$

% change in de Broglie wavelength$= \left( \frac{\lambda - \lambda'}{\lambda} \right) \times 100$

$= \left( 1 - \frac{\lambda'}{\lambda} \right)100 = \left( 1 - \frac{1}{4} \right) \times 100 = 75\%$