Question

An electron, an $\alpha$- particle and a proton have the same K.E. Which of these particles has the shortest de Broglie wavelength?

• A. electron
• B. proton
• C. $\alpha$- particle
• D. electron, proton

Answer 1

The correct option is C $\alpha$- particle

$K_E=\frac{1}{2}m{v}^2=\frac{p^2}{2m}p=\sqrt{2mE}\therefore \lambda =\frac{h}{\sqrt{2mE}}\Rightarrow \lambda \alpha \frac{1}{\sqrt{m}}{m}_p=1836m_e{m}_{\alpha }=4\times 1836m_{e}therefore{m}_{\alpha }>m_p>m_e$
$\Rightarrow {\lambda }_{\alpha }$ is the shortest