The energy of an electron in an orbit of hydrogen like ion with an orbit radius of \(52.9 \mathrm{pm}\) in…
The energy of an electron in an orbit of hydrogen like ion with an orbit radius of \(52.9 \mathrm{pm}\) in \(\mathrm{J}\) is (ground state energy of electron in hydrogen atom is \(=-2.18 \times 10^{-18} \mathrm{~J}\) )
\(-4.36 \times 10^{-18}\)
\(-1.09 \times 10^{-17}\)
\(-8.72 \times 10^{-18}\)
\(-6.54 \times 10^{-18}\)
Solution
Given,
Orbit radius \(=52.9 \mathrm{pm}\)
Ground state energy of electron in hydrogen atom
\(\begin{aligned}
& =-2.18 \times 10^{-18} \mathrm{~J} \\
r_n & =r_0 \times \frac{n^2}{Z} \\
52.9 \mathrm{pm} & =52.9 \mathrm{pm} \times \frac{n^2}{Z} \Rightarrow \frac{n^2}{Z}=1 \text { or } n^2=Z
\end{aligned}\)
From energy,
\(\begin{aligned}
& E_n=E_0 \times \frac{Z^2}{n^2} \\
& \Rightarrow \quad E_n=-2.18 \times 10^{-18} \times \frac{n^4}{n^2} \\
& E_n=-2.18 \times 10^{-18} \times n^2 \\
& \left(\because n^2=Z\right) \\
& E_n=4 \times-2.18 \times 10^{-18}=-8.72 \times 10^{-18} \mathrm{~J} \\
\end{aligned}\)
If \(n=2\), from given option, then
\(E_n=4 \times-2.18 \times 10^{-18}=-8.72 \times 10^{-18} \mathrm{~J}\)
Thus, option (3) is correct.