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, $p_{\alpha \text {-particle }}=\frac{1}{2} p_{\text {proton }}$
$\Rightarrow \frac{p_{\alpha \text {-particle }}}{p_{\text {proton }}}=\frac{1}{2} \ldots \text { (i) }$
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)]