Question

If the kinetic energy of the particle is increased to $16$ times the initial value, the percentage change in the de-Broglie wavelength of the particle is,

• A. $25\%$
• B. $75\%$
• C. $60\%$
• D. $50\%$

Answer 1

The correct option is B $75\%$

The de-Broglie wavelength is given by,
$\lambda =\frac{h}{\sqrt{2m(K.E)}}$
$\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{(K.E{)}_2}{(K.E{)}_1}}=\sqrt{\frac{16(K.E)}{(K.E)}}=4$
$\Rightarrow {\lambda }_2=\frac{{\lambda }_1}{4}$

$\therefore$ The percentage change in $\lambda$
$=\frac{{\lambda }_2-{\lambda }_1}{{\lambda }_1}\times 100=\frac{\frac{{\lambda }_1}{4}-{\lambda }_1}{{\lambda }_1}\times 100$

$=-75\%$

Negative sign indicates that $\lambda$ decreases.

Hence, $\left(B\right)$ is the correct answer.