Question

If the momentum of an $\alpha$-particle is half that of a proton, then the ratio between the wavelengths of their deBroglie waves is

• A. $1:2$
• B. $4:1$
• C. $1:4$
• D. $1:1$
• E. $2:1$

Answer 1

Answer: (E)

Given, $p_{\alpha \text{-particle }}=\frac{1}{2}{p}_{\text{proton }}$
$\Rightarrow \frac{p_{\alpha \text{-particle }}}{p_{\text{proton }}}=\frac{1}{2}\dots \text{ (i) }$
de-Broglie wavelength is given as, $\lambda =h/p$
where, $h$ is Planck's constant and $p$ is momentum.
$\Rightarrow \lambda \propto \frac{1}{p}(\because p\text{ is constant })$
$\Rightarrow \frac{{\lambda }_{\alpha \text{-particle }}}{{\lambda }_{\text{proton }}}=\frac{p_{\text{proton }}}{p_{\alpha -\text{ particle }}}$
$=\frac{2}{1}$
[from Eq. (i)]