The energy and radius of electron present in second orbit of $\mathrm{He}^{+}$respectively are

The energy and radius of electron present in second orbit of $\mathrm{He}^{+}$respectively are
  1. $-1.09 \times 10^{-18} \mathrm{~J}, 105.8 \mathrm{pm}$
  2. $-8.72 \times 10^{-18} \mathrm{~J}, 211.6 \mathrm{pm}$
  3. $-4.36 \times 10^{-18} \mathrm{~J}, 52.9 \mathrm{pm}$
  4. $-2.18 \times 10^{-18} \mathrm{~J}, 105.8 \mathrm{pm}$

Solution

$\begin{aligned} & \text { For } \mathrm{He}^{+}: Z=2, n=2 \\ & E_n=-2.18 \times 10^{-18} \cdot \frac{Z^2}{n^2} \cdot \mathrm{J} \\ & \therefore \quad E_n=-2.18 \times 10^{-18} \times \frac{4}{4} \\ & =-2.18 \times 10^{-18} \mathrm{~J} \\ & r_n=52.9 \times \frac{n^2}{Z} \mathrm{pm}=52.9 \times \frac{4}{2}=105.8 \mathrm{pm} \\ & \end{aligned}$

Asked in: AP EAMCET 2017 (26 Apr Shift 1)

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