An electron, a doubly ionized helium ion ($He^{++}$) and proton are having the same kinetic energy. The…

An electron, a doubly ionized helium ion ($He^{++}$) and proton are having the same kinetic energy. The relation between their respective de-Broglie wavelength λe*λHe++ and λp is :
  1. $\lambda_e>\lambda_{{He}^{++}}>\lambda_p$
  2. $\lambda_e<\lambda_{{He}^{++}}==\lambda_p$
  3. $\lambda_e>\lambda_p>\lambda_{{He}^{++}}$
  4. $\lambda_e<\lambda_p<\lambda_{{He}^{++}}$

Solution

$\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 m k}}$
$\lambda_{\mathrm{e}}: \lambda_{\mathrm{p}}: \lambda_{{He}^{++}}=\frac{1}{\sqrt{\mathrm{m}_{\mathrm{e}}}}: \frac{1}{\sqrt{\mathrm{m}_{\mathrm{p}}}}: \frac{1}{\sqrt{\mathrm{He}^{++}}}$
$\frac{1}{\sqrt{9.1 \times 10^{-31}}}: \frac{1}{\sqrt{1.6 \times 10^{-27}}}: \frac{1}{\sqrt{4 \times 1.6 \times 10^{-27}}}=84: 2: 1$

Asked in: JEE Main 2020 (06 Sep Shift 1)

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