Calculate the energy of an electron in the first four orbits of a hydrogen atom in $kJ \, mol^{-1}$ (Given: $N_A = 6.02 \times 10^{23} \, mol^{-1}$,Rydberg constant $R_H = 2.18 \times 10^{-18} \, J$).

(N/A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = -\frac{R_H}{n^2} \, J \, \text{atom}^{-1}$.
To convert this to $kJ \, mol^{-1}$,we multiply by Avogadro's number $(N_A)$ and divide by $1000$:
$E_n (kJ \, mol^{-1}) = -\frac{2.18 \times 10^{-18} \times 6.02 \times 10^{23}}{n^2 \times 1000} \approx -\frac{1312}{n^2} \, kJ \, mol^{-1}$.

Orbit $(n)$	Energy $(kJ \, mol^{-1})$
$1$	$-1312$
$2$	$-328$
$3$	$-145.8$
$4$	$-82$