Q. If the kinetic energy of the 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%
Explanation
(b) $\lambda = \frac{h}{\sqrt{2mK}} \Rightarrow \frac{\lambda_{1}}{\lambda_{2}} = \sqrt{\frac{K_{2}}{K_{1}}} = \sqrt{\frac{16K}{K}} = 4$
$\frac{\lambda_{1}}{\lambda_{2}} = 4 \Rightarrow \lambda_{2} = \frac{\lambda_{1}}{4} = \frac{100}{4} = 25$
$4\lambda = 100 - 25 = 75\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.