Let the shortest wavelength of the Lyman series of a hydrogen atom be . If the wavelength of the fir — Atomic Structure Chemistry Question

💡 Solution & Explanation

The shortest wavelength of the Lyman series occurs when $n_2=\infty$, so $\frac{1}{x} = R_H(\frac{1}{1^2} - 0) = R_H$, hence $x = \frac{1}{R_H}$. The first line of the Balmer series ($n_1=2, n_2=3$) has $\frac{1}{\lambda} = R_H(\frac{1}{2^2} - \frac{1}{3^2}) = R_H(\frac{1}{4} - \frac{1}{9}) = \frac{5 R_H}{36}$. Therefore, $\lambda = \frac{36}{5 R_H} = \frac{36x}{5}$. Thus, $k = 36$.