Question

If the kinetic energy of a 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)

Let the initial de Broglie wavelength be ${\lambda }_0=\frac{h}{m{v}_0}$

The kinetic energy is increased 16 times.

Thus $K{E}^{\prime }=16K{E}_0$

$⟹\frac{1}{2}m{v}^2=16\times \frac{1}{2}m{v}_0^2$

$⟹v=4v_0$

Thus the new de Broglie wavelength=$\lambda =\frac{h}{mv}=\frac{h}{m\left(4v_0\right)}=\frac{{\lambda }_0}{4}$

Thus the percentage change in wavelength: $=\frac{{\lambda }_0-\lambda }{{\lambda }_0}\times 100=75$%