The energy of the orbit of hydrogen is $E_n = \frac{1.31 \times 10^6}{n^2} \ J \ mol^{-1}$. If an electron transits from $n = 3$ to $n = 2$,find the frequency of the emitted radiation. (Note: $h = 6.6 \times 10^{-34} \ J \ s$,$N_A = 6.02 \times 10^{23} \ mol^{-1}$)

(A) The energy change $\Delta E$ for the transition is given by $\Delta E = E_3 - E_2 = 1.31 \times 10^6 \times (\frac{1}{2^2} - \frac{1}{3^2}) \ J \ mol^{-1}$.
$\Delta E = 1.31 \times 10^6 \times (\frac{1}{4} - \frac{1}{9}) = 1.31 \times 10^6 \times \frac{5}{36} \ J \ mol^{-1}$.
$\Delta E = 1.8194 \times 10^5 \ J \ mol^{-1}$.
To find the energy per atom,divide by Avogadro's number: $E_{atom} = \frac{1.8194 \times 10^5}{6.02 \times 10^{23}} \ J \approx 3.022 \times 10^{-19} \ J$.
Using the relation $E = h\nu$,the frequency $\nu = \frac{E}{h} = \frac{3.022 \times 10^{-19}}{6.6 \times 10^{-34}} \ Hz$.
$\nu \approx 4.58 \times 10^{14} \ Hz$.