Question

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%

Answer 1

Answer: (B)

From de-Broglies formula $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mK}}$ $\frac{{\lambda }_1}{{\lambda }_2}=\frac{\sqrt{K_2}}{\sqrt{K_1}}$ Given, $K_2=16K_1$ $\therefore$ $\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{16}=4$ $\Rightarrow$ ${\lambda }_2=\frac{{\lambda }_1}{4}$ % change $=\frac{{\lambda }_2-{\lambda }_1}{{\lambda }_1}\times 100$ % change $=\frac{3}{4}\times 100=75$