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)

de Broglie wavelength,
$\lambda =\frac{h}{\sqrt{2mK}}$
where $m$ is the mass and $K$ is the kinetic energy of the particle.
When kinetic energy of the particle is increased to 16 times, then its de Broglie wavelength becomes,
${\lambda }^{\prime }=\frac{h}{\sqrt{2m(16K)}}=\frac{1}{4}\frac{\lambda }{\sqrt{2mK}}=\frac{\lambda }{4}$( Using ( i ))
$\%$ change in the de Broglie wavelength
$=\frac{\lambda -{\lambda }^{\prime }}{\lambda }\times 100=\left(1-\frac{{\lambda }^{\prime }}{\lambda }\right)\times 100$
$=\left(1-\frac{1}{4}\right)\times 100=75\%$