Question

If an alpha particle, a proton and an electron are moving with the same momentum, then their respective de-Broglie wavelengths ${\lambda }_{\alpha },{\lambda }_p,\text{and}{\lambda }_e$ will be related as,

• A. ${\lambda }_{\alpha }={\lambda }_p={\lambda }_e$
• B. ${\lambda }_{\alpha }<{\lambda }_p<{\lambda }_e$
• C. ${\lambda }_{\alpha }>{\lambda }_p>{\lambda }_e$
• D. ${\lambda }_p>{\lambda }_e>{\lambda }_{\alpha }$

Answer 1

The correct option is A ${\lambda }_{\alpha }={\lambda }_p={\lambda }_e$

The de-Broglie wavelength is given as
$\lambda =\frac{h}{p}$

$\because p_{\alpha }=p_p=p_e$

$\therefore {\lambda }_{\alpha }={\lambda }_p={\lambda }_e$

Hence, $\left(A\right)$ is the correct answer.