Question

If the kinetic energy of the particle is increased to $16$ times its previous value, 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

$\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 m E}}$ or $\lambda \propto \frac{1}{\sqrt{E}}$
$\therefore \frac{\lambda^{\prime}}{\lambda}=\sqrt{\frac{E}{E'}}=\sqrt{\frac{1}{16}}=\frac{1}{4}$
$\%$ change in $=\left(\frac{\lambda-\lambda'}{\lambda}\right) \times 100$
$=\left(1-\frac{\lambda^{\prime}}{\lambda}\right) \times 100=\left(1-\frac{1}{4}\right) \times 100=75 \%$