Question

Masses of two particles having same kinetic energies are in the ratio 1 : 2 then, their de-broglie wavelengths are in ratio:

• A. $2:1$
• B. $1:2$
• C. $\sqrt{2}:1$
• D. $\sqrt{3}:1$

Answer 1

The correct option is C

$\sqrt{2}:1$

$\frac{m_1}{m_2}=\frac{1}{2}$
Two particles having same kinetic energies
$k_1=k_2\frac{1}{2}{m}_1{v}_1^2=\frac{1}{2}{m}_2{v}_2^2\frac{m_1}{m_2}=[\frac{v_2}{v_1}{]}^2\frac{1}{\sqrt{2}}=\frac{v_2}{v_1}{v}_1=\sqrt{2}{v}_2\lambda \infty \frac{1}{mv}\frac{{\lambda }_1}{{\lambda }_2}=\frac{m_2{v}_2}{m_1{v}_1}\frac{{\lambda }_1}{{\lambda }_2}=\frac{2v_2}{\sqrt{2}{v}_2}=\frac{\sqrt{2}}{1}$