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 $n _{2}=2,3,4, \ldots \ldots \infty$
$\frac{1}{\lambda}= R \left(1-\frac{1}{ n _{2}^{2}}\right) $
$\lambda=\lambda_{\max }, $ when $ n _{2}=2$
$\frac{1}{\lambda_{\max }}=\frac{3 R }{4} $
$\Rightarrow \lambda_{\max }=\frac{4}{3 R }=121.5\, nm $
$\lambda=\lambda_{\min },$ when $n _{2}=\infty $
$\frac{1}{\lambda_{\min }}= R $
$\Rightarrow \lambda_{\min }=\frac{1}{ R }=91.1\, nm$
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 $n _{2}=3$
$\frac{1}{\lambda_{\max }}=\frac{5 R }{36}$
$\lambda_{\max }=\frac{36}{5 R }=656.2\, nm$
$\lambda=\lambda_{\min }$, when $n _{2}=\infty$
$\Rightarrow \lambda_{\min }=\frac{4}{ R }=364.5\, nm$
Hence, for the Balmer series,
$\frac{\lambda_{\max }}{\lambda_{\min }}=\frac{36 / 5 R }{4 / R }=\frac{9}{5}$
For the Paschen series, $n _{1}=3$ and $n _{2}=4,5,6, \ldots \ldots ., \infty$
$\frac{1}{\lambda}= R \left(\frac{1}{9}-\frac{1}{ n _{2}^{2}}\right)$
$\lambda=\lambda_{\max }$, when $n _{2}=4$
$\frac{1}{\lambda_{\max }}= R \left(\frac{1}{9}-\frac{1}{16}\right)=\frac{7 R }{144}$
$\Rightarrow \lambda_{\max }=\frac{144}{7 R }=1874.7 \,nm$
$\frac{1}{\lambda_{\max }}= R \left(\frac{1}{9}-\frac{1}{16}\right)=\frac{7 R }{144}$
$\Rightarrow \lambda_{\max }=\frac{144}{7 R }=1874.7\, nm$
$\lambda=\lambda_{\min }$, when $n _{2}=\infty$
$\frac{1}{\lambda_{\min }}=\frac{ R }{9}$
$ \Rightarrow \lambda_{\min }=\frac{9}{ R }=820.2\, nm$