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)

$4E$

Answer 2

Since Bohr?? formula for energy in $n^{th}$ orbit is $E_{n} = -\frac{me^{4}Z^{2}}{8\varepsilon_{0}^{2}n^{2}h^{2}}$ $\Rightarrow E_{n} \propto Z^{2}$ Now for hydrogen $\left(_{1}H^{2}\right), Z_{H}= 1,$and $E_H = E$ and for helium $\left(_{2}H^{4}\right), Z_{He}= 2$, $\therefore \frac{E_{H}}{E_{He}} = \frac{\left(Z_{H}\right)^{2}}{\left(Z_{He}\right)^{2}}$ $\Rightarrow E_{He} = \left(\frac{Z_{He}}{Z_{H}}\right)^{2} E_{H}$ $= \left(2\right)^{2} E = 4 E$