Question

Two particles of masses $m$ and $2m$ have equal kinetic energies. Their de Broglie wavelengths are in the ratio of :

• A. $1:1$
• B. $1:2$
• C. $1:\sqrt{2}$
• D. $\sqrt{2}:1$

Answer 1

Answer: (D)

$\sqrt{2}:1$

de Broglie wavelength is $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mk}}$ where $k=$ kinetic energy

$\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\frac{\sqrt{m_2{k}_2}}{\sqrt{m_1{k}_1}}=\sqrt{\frac{m_2}{m_1}}=\frac{\sqrt{2}}{1}$

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{\sqrt{2}}{1}$