Question

The energy of an electron in an orbit of hydrogen like ion with an orbit radius of $52.9\, pm$ in $J$ is
(ground state energy of electron in hydrogen atom is $ - 2.18 \times 10^{-18} J$)

• A. $ - 4.36 \times 10^{-18}$
• B. $ - 1.09 \times 10^{-17}$
• C. $ - 8.72 \times 10^{-18}$
• D. $ - 6.54 \times 10^{-18}$

Answer 1

Answer: (C)

Given,

Orbit radius $=52.9 \,pm$

Ground state energy of electron in hydrogen atom

$=-2.18 \times 10^{-18}\, J$

$r_{n} =r_{0} \times \frac{n^{2}}{Z} $

$52.9\, pm=52.9 \,pm \times \frac{n^{2}}{Z}$

$ \Rightarrow \frac{n^{2}}{Z}=1$ or $n^{2}=Z$

From energy,

$E_{n} =E_{0} \times \frac{Z^{2}}{n^{2}} $

$\Rightarrow E_{n} =-2.18 \times 10^{-18} \times \frac{n^{4}}{n^{2}} $

$E_{n} =-2.18 \times 10^{-18} \times n^{2}$

$\left(\because n^{2}=Z\right)$

$E_{n}=-2.18 \times 10^{-18} \times n^{2}$

If $n=2$, from given option, then

$E_{n}=4 \times-2.18 \times 10^{-18}$

$=-8.72 \times 10^{-18} \,J$