If the radius of electron orbit in the excited state of hydrogen atom is $476.1 \mathrm{pm}$, the energy of…

If the radius of electron orbit in the excited state of hydrogen atom is $476.1 \mathrm{pm}$, the energy of electron in that excited state in $\mathrm{J}$ is (Radius and energy of electron in the first orbit of hydrogen atom are $52.9 \mathrm{pm}$ and $-2.18 \times 10^{-18} \mathrm{~J}$ respectively)
  1. $-2.42 \times 10^{-18}$
  2. $-19.62 \times 10^{-18}$
  3. $-2.42 \times 10^{-19}$
  4. $-6.05 \times 10^{-19}$

Solution

Given : Radius of excited state of hydrogen $ \begin{aligned} & \text { atom }=476 \cdot 1 \mathrm{PM} \\ & \qquad \begin{aligned} E_1 & =2.18 \times 10^{-8} \mathrm{~J} / \text { atom } \\ E_n & =-2.18 \times 10^{-18} \cdot \frac{Z^2}{n^2} \\ r_1 & =529 \mathrm{pm} \\ r_n & =\frac{n^2 \times 527}{Z} ; 476.1=\frac{n^2 \times 52.7}{Z} \\ \therefore n^2=9 & \text { and } E_n=E_3=-\frac{2.18 Z^2 \times 10^{-18} \mathrm{~J}}{9} \\ & =-2.42 \times 10^{-19} \mathrm{~J} \end{aligned} \end{aligned} $

Asked in: AP EAMCET 2018 (22 Apr Shift 1)

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