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 $9/5$
• B. There is an overlap between the wavelength ranges of Balmer and Paschen series
• C. The wavelength of Lyman series are given by $\left(1+\frac{1}{m^2}\right){\lambda }_0$, 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)

For hydrogen atom, $z=1$
$\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$,
where $R=1.0973\times 10^7{m}^{-1}$
$=1.1\times 10^7{m}^{-1}=$ Rydberg constant.
For the Lyman series, $n_1=1$ and $n_2=2,3,4,\dots \dots \infty$
$\frac{1}{\lambda }=R\left(1-\frac{1}{n_2^2}\right)$
$\lambda ={\lambda }_{\max },$ when $n_2=2$
$\frac{1}{{\lambda }_{\max }}=\frac{3R}{4}$
$\Rightarrow {\lambda }_{\max }=\frac{4}{3R}=121.5nm$
$\lambda ={\lambda }_{\min },$ when $n_2=\infty$
$\frac{1}{{\lambda }_{\min }}=R$
$\Rightarrow {\lambda }_{\min }=\frac{1}{R}=91.1nm$
Also, $\lambda =\left(\frac{n_2^2}{n_2^2-1}\right)\frac{1}{R}=\left[1+\frac{1}{\left(n_2^2-1\right)}\right]{\lambda }_0$
$=\left(1+\frac{1}{m^2}\right){\lambda }_0$
$\lambda =\left(1+\frac{1}{m^2}\right){\lambda }_0$
where,
$m^2=\left(n_2^2-1\right)=$ an integer
$m=\sqrt{n_2^2-1}=$ not an integer
For the Balmer series, $n_1=2$ and $n_2=3,4,5,6,........,\infty$
$\frac{1}{\lambda }=R\left(\frac{1}{4}-\frac{1}{n_2^2}\right)$
$\lambda ={\lambda }_{\max }$, when $n_2=3$
$\frac{1}{{\lambda }_{\max }}=\frac{5R}{36}$
${\lambda }_{\max }=\frac{36}{5R}=656.2nm$
$\lambda ={\lambda }_{\min }$, when $n_2=\infty$
$\Rightarrow {\lambda }_{\min }=\frac{4}{R}=364.5nm$
Hence, for the Balmer series,
$\frac{{\lambda }_{\max }}{{\lambda }_{\min }}=\frac{36/5R}{4/R}=\frac{9}{5}$
For the Paschen series, $n_1=3$ and $n_2=4,5,6,\dots \dots .,\infty$
$\frac{1}{\lambda }=R\left(\frac{1}{9}-\frac{1}{n_2^2}\right)$
$\lambda ={\lambda }_{\max }$, when $n_2=4$
$\frac{1}{{\lambda }_{\max }}=R\left(\frac{1}{9}-\frac{1}{16}\right)=\frac{7R}{144}$
$\Rightarrow {\lambda }_{\max }=\frac{144}{7R}=1874.7nm$
$\frac{1}{{\lambda }_{\max }}=R\left(\frac{1}{9}-\frac{1}{16}\right)=\frac{7R}{144}$
$\Rightarrow {\lambda }_{\max }=\frac{144}{7R}=1874.7nm$
$\lambda ={\lambda }_{\min }$, when $n_2=\infty$
$\frac{1}{{\lambda }_{\min }}=\frac{R}{9}$
$\Rightarrow {\lambda }_{\min }=\frac{9}{R}=820.2nm$