Q1 / 160 · AP EAMCET 2008 · Chemistry
An electronic transition in hydrogen atom results in the formation of $\mathrm{H}_\alpha$ line of hydrogen in Lyman series, the energies associated with the electron in each of the orbits involved in the transition (in kcal $\mathrm{mol}^{-1}$ ) are
- A$-313.6,-34.84$
- Correct answerB$-313.6,-78.4$
- C$-78.4,-34.84$
- D$-78.4,-19.6$
Solution
Energy of an electron in $n^{\text {th }}$ orbit,
$$
E_n=-\frac{2 \pi^2 k^2 m Z^2 e^4}{n^2 h^2}
$$
On substituting the values of $k, m, e$ and $h$, we get
$$
\begin{aligned}
E_n & =-\frac{2.172 \times 10^{-18} Z^2}{m^2} \mathrm{~J}^2 \text { atom }^{-1} \\
\text { or } \quad & =-\frac{1311.8 Z^2}{n^2} \mathrm{~kJ} \mathrm{~mol}^{-1} \\
\text { or } \quad & =-\frac{313.52 Z^2}{n^2} \mathrm{kcal} \mathrm{mol}^{-1} \\
& {[\because 1 \mathrm{kcal}=4.184 \mathrm{~kJ}] }
\end{aligned}
$$
For H-atom, $\quad Z=1$
For Lyman series, $n_1=1, n_2=2$
Energy of electron in $n_1$ orbit
$$
\begin{aligned}
& =-\frac{313.52 \times(1)^2}{(1)^2} \mathrm{kcal} \mathrm{mol}^{-1} \\
& =-313.52 \mathrm{kcal} \mathrm{mol}^{-1} \\
& \approx-313.6 \mathrm{kcal} \mathrm{mol}^{-1}
\end{aligned}
$$
Energy of electron in $n_2$ orbit
$$
\begin{aligned}
& =-\frac{313.52 \times(1)^2}{(2)^2} \mathrm{kcal} \mathrm{mol}^{-1} \\
& =-\frac{313.52}{4} \mathrm{kcal} \mathrm{mol}^{-1} \\
& =-78.38 \mathrm{kcal} \mathrm{mol}^{-1}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.