In the third orbit of hydrogen atom the energy of an electron ' $E$ '. In the fifth orbit of helium $(Z=2)$…
In the third orbit of hydrogen atom the energy of an electron ' $E$ '. In the fifth orbit of helium $(Z=2)$ the energy of an electron will be
- $\frac{25 \mathrm{E}}{36}$
- $\frac{36 \mathrm{E}}{25}$
- $\frac{3 \mathrm{E}}{5}$
- $\frac{5 \mathrm{E}}{3}$
Solution
$\mathrm{E} \propto \frac{\mathrm{Z}^2}{\mathrm{n}^2}$
$\begin{aligned} \therefore \quad \frac{E_{\mathrm{H}}}{\mathrm{E}_{\mathrm{He}}} & =\frac{\left(\mathrm{Z}^2\right)_{\mathrm{H}}}{(\mathrm{n})_{\mathrm{H}}^2} \times \frac{(\mathrm{n})_{\mathrm{He}}^2}{\left(\mathrm{Z}^2\right)_{\mathrm{He}}} \\ & =\frac{1}{3^2} \times \frac{5^2}{2^2}=\frac{25}{36} \\ \therefore \quad \mathrm{E}_{\mathrm{He}} & =\frac{36}{25} \mathrm{E}_{\mathrm{H}}=\frac{36}{25} \mathrm{E}\end{aligned}$
Asked in: MHT CET 2024 (02 May Shift 1)
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