Question

An electron, a doubly ionized helium ion (${\mathrm{He}}^{2+}$) and a proton are having the same kinetic energy. The relation between their respective de-Broglie wavelengths ${\mathrm{\lambda }}_{\mathrm{e}}$, ${\mathrm{\lambda }}_{{\mathrm{He}}^{2+}}$ and ${\mathrm{\lambda }}_{\mathrm{p}}$ is

• A. ${\mathrm{\lambda }}_{\mathrm{e}}$ $>$ ${\mathrm{\lambda }}_{\mathrm{p}}$ $>$ ${\mathrm{\lambda }}_{{\mathrm{He}}^{++}}$
• B. ${\mathrm{\lambda }}_{\mathrm{e}}$ $>$ ${\mathrm{\lambda }}_{{\mathrm{He}}^{++}}$$>$ ${\mathrm{\lambda }}_{\mathrm{p}}$
• C. ${\mathrm{\lambda }}_{\mathrm{e}}$ $<$ ${\mathrm{\lambda }}_{\mathrm{p}}$ $<$ ${\mathrm{\lambda }}_{{\mathrm{He}}^{++}}$
• D. ${\mathrm{\lambda }}_{\mathrm{e}}$ $<$ ${\mathrm{\lambda }}_{{\mathrm{He}}^{++}}$ $=$ ${\mathrm{\lambda }}_{\mathrm{p}}$

Answer 1

The correct option is A

${\mathrm{\lambda }}_{\mathrm{e}}$ $>$ ${\mathrm{\lambda }}_{\mathrm{p}}$ $>$ ${\mathrm{\lambda }}_{{\mathrm{He}}^{++}}$

Explanation for the correct option:

We know,

De Broglie’s wavelength $\lambda =\frac{h}{mv}=\frac{h}{p}$

$KE=\frac{1}{2}m{v}^2=\frac{{\left(mv\right)}^2}{2m}=\frac{p^2}{2m}$, $P=\sqrt{2mKE}$

So, $\lambda =\frac{h}{\sqrt{2mKE}}$

$\mathrm{\lambda }$ $\propto$$\frac{1}{\sqrt{m}}$ $\Rightarrow$ $\lambda =\frac{C}{\sqrt{m}}$

Here m is mass.

The sequence of mass of the particles is ${\mathrm{m}}_{{\mathrm{He}}^{++}}>{\mathrm{m}}_{\mathrm{p}}>{\mathrm{m}}_{\mathrm{e}}$

Therefore, ${\mathrm{\lambda }}_{\mathrm{e}}$ $>$ ${\mathrm{\lambda }}_{\mathrm{p}}$ $>$ ${\mathrm{\lambda }}_{{\mathrm{He}}^{++}}$

Hence option (a) is correct.