Question

Which of the following statement(s) is(are) correct about the spectrum of hydrogen atom ?

• A. The ratio of the longest wavelength to the shortest wavelength in Balmer series is $\frac{9}{5}$
• B. There is an overlap between the wavelength ranges of Balmer and Paschen series
• C. The wavelengths of Lyman series are given by $(1+\frac{1}{m^2}){\lambda }_0$, where ${\lambda }_0$ is the shortest wavelength of Lyman series and $m$ is an integer.
• D. The wavelength ranges of Lyman and Balmer series do not overlap

Answer 1

Answer: (A), (D)

The wavelength ranges of Lyman and Balmer series do not overlap

[A]
When the transition is from any level to $n=2$, then photon emitted belongs to Balmer series.
Therefore, for longest wavelength, transition occurs from $n=3$ to $n=2$ and for shortest wavelength transition occurs from $n=\infty$ to $n=2$.
$\therefore \frac{1}{\lambda }=R{z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$\frac{{\lambda }_{longest}}{{\lambda }_{shortest}}=\frac{[\frac{1}{2^2}-\frac{1}{\infty }]}{[\frac{1}{2^2}-\frac{1}{3^2}]}=\frac{9}{5}$
[B]
As we know that,
${\lambda }_{longest}$ of Balmer $=\frac{36}{5R}$
${\lambda }_{shortest}$ of Paschen $\frac{9}{R}$, hence, they don't overlap as the transition is from $n=\infty$ to $n=3$.
[C]
For Lymen series
$\frac{1}{\lambda }=R[\frac{1}{1}-\frac{1}{m^2}]$ also, $\frac{1}{{\lambda }_0}=R$
$\therefore \frac{1}{\lambda }=\frac{1}{{\lambda }_0}[1-\frac{1}{m^2}]\Rightarrow \lambda =\frac{{\lambda }_0}{1-\frac{1}{m^2}}$
[D]
${\lambda }_{longest}$ of Lymen $=\frac{4}{3R}$, ${\lambda }_{shortest}$ of Balmer $=\frac{4}{R}$, hence, these wavelength don't overlap.