Question

If $E$ is the energy of $n^{th}$ orbit of hydrogen atom the energy of $n^{th}$ orbit of $He$ atom will be

• A. $E$
• B. $2E$
• C. $3E$
• D. $4E$

Answer 1

Answer: (D)

Since Bohr’s formula for energy in $n^{th}$ orbit is
$E_n=-\frac{m{e}^4{Z}^2}{8{\epsilon }_0^2{n}^2{h}^2}$
$\Rightarrow E_n\propto Z^2$
Now for hydrogen $\left({}_1{H}^2\right),Z_H=1,$and $E_H=E$ and for helium $\left({}_2{H}^4\right),Z_{He}=2$,
$\therefore \frac{E_H}{E_{He}}=\frac{{\left(Z_H\right)}^2}{{\left(Z_{He}\right)}^2}$
$\Rightarrow E_{He}={\left(\frac{Z_{He}}{Z_H}\right)}^2{E}_H$
$={\left(2\right)}^{2}E=4E$