Question

The ratio of de Broglie wavelength of an $\alpha$- particle and a proton of same kinetic energy is

• A. 1:2
• B. 1:1
• C. $1:\sqrt{2}$
• D. 4:1

Answer 1

The correct option is A 1:2

$\lambda =\frac{h}{\sqrt{2m{E}_k}}\frac{{\lambda }_{\alpha }}{{\lambda }_p}=\sqrt{\frac{m_p}{m_{\alpha }}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$