MHT CET · Physics · Atomic Physics
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
- A \(\frac{25 \mathrm{E}}{36}\)
- B \(\frac{36 \mathrm{E}}{25}\)
- C \(\frac{3 \mathrm{E}}{5}\)
- D \(\frac{5 \mathrm{E}}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{36 \mathrm{E}}{25}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{E} \propto \frac{\mathrm{Z}^2}{\mathrm{n}^2}\)
\(\begin{aligned} \therefore \quad \frac{E_{\mathrm{H}}}{\mathrm{E}_{\mathrm{He}}} & =\frac{\left(\mathrm{Z}^2\right)_{\mathrm{H}}}{(\mathrm{n})_{\mathrm{H}}^2} \times \frac{(\mathrm{n})_{\mathrm{He}}^2}{\left(\mathrm{Z}^2\right)_{\mathrm{He}}} \\ & =\frac{1}{3^2} \times \frac{5^2}{2^2}=\frac{25}{36} \\ \therefore \quad \mathrm{E}_{\mathrm{He}} & =\frac{36}{25} \mathrm{E}_{\mathrm{H}}=\frac{36}{25} \mathrm{E}\end{aligned}\)
\(\begin{aligned} \therefore \quad \frac{E_{\mathrm{H}}}{\mathrm{E}_{\mathrm{He}}} & =\frac{\left(\mathrm{Z}^2\right)_{\mathrm{H}}}{(\mathrm{n})_{\mathrm{H}}^2} \times \frac{(\mathrm{n})_{\mathrm{He}}^2}{\left(\mathrm{Z}^2\right)_{\mathrm{He}}} \\ & =\frac{1}{3^2} \times \frac{5^2}{2^2}=\frac{25}{36} \\ \therefore \quad \mathrm{E}_{\mathrm{He}} & =\frac{36}{25} \mathrm{E}_{\mathrm{H}}=\frac{36}{25} \mathrm{E}\end{aligned}\)
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