Question

The energy of an electron in an orbit of hydrogen like ion with an orbit radius of $52.9pm$ 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 $=52.9pm$

Ground state energy of electron in hydrogen atom

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

$r_n=r_0\times \frac{n^2}{Z}$

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

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

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

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