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. $60$
• C. $50$
• D. $75$

Answer 1

The correct option is D $75$

Relation between $\lambda$ and kinetic energy is
$\lambda =\frac{h}{\sqrt{2mK}}$

Given that $K_2=16K,$ let wavelength be ${\lambda }_2$

Ratio of wavelengths is
$\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{K_2}{K_1}}=\sqrt{\frac{16K}{K}}=\frac{4}{1}$
Percentage change in wavelength is $\frac{{\lambda }_2-{\lambda }_1}{{\lambda }_1}\times 100=\frac{1-4}{4}\times 100=-75\text{\%}$