JEE Advance - Physics (2021 - Paper 1 Online - No. 14)
Which of the following statement(s) is(are) correct about the spectrum of the hydrogen atom?
The ratio of the longest wavelength to the shortest wavelength in Balmer series is 9/5
There is an overlap between the wavelength ranges of Balmer and Paschen series
The wavelengths of Lyman series are given by $$\left( {1 + {1 \over {{m^2}}}} \right){\lambda _0}$$, where $${\lambda _0}$$ is the shortest wavelength of Lyman series and m is an integer
The wavelength ranges of Lyman and Balmer series do not overlap
Explanation
For Balmer series
$$ \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{n^2}\right) n>2 $$
$\lambda$ is minimum for transition from $n_1 \rightarrow \infty$ to $n_2=2$.
$$ \frac{1}{\lambda_{\min }}=\frac{R}{4} $$
$\lambda$ is maximum for transition from $n_1=3$ to $n_2=2$
$$ \frac{1}{\lambda_{\max }}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5 \mathrm{R}}{36} $$
Dividing the above two equations, we get
$$ \frac{\lambda_{\max }}{\lambda_{\min }}=\frac{R}{4} \times \frac{36}{5 R}=\frac{9}{5} $$
Option (A) is correct.
For Lyman series,
$\lambda_{\min }=91.18 \mathrm{~nm}$
$\lambda_{\max }=121.57 \mathrm{~nm}$
For Balmer series,
$\lambda_{\min }=364.60 \mathrm{~nm}$
$\lambda_{\max }=656.30 \mathrm{~nm}$
For Pashen series,
$\lambda_{\min }=820.40 \mathrm{~nm}$
$\lambda_{\max }=1875 \mathrm{~nm}$
We can see from the above that there is no overlap between the wavelength ranges of Balmer and Paschen series. so, option (B) is incorrect.
Similarly wavelength ranges of Lyman and Balmer series also do not overlap. Hence, option (D) is correct.
Shortest wavelength of Lyman series
$$ \begin{aligned} \frac{1}{\lambda_0} & =R\left(1-\frac{1}{\infty}\right) \\\\ \lambda_0 & =\frac{1}{R} \end{aligned} $$
For Lyman series, $\frac{1}{\lambda}=\mathrm{R}\left(1-\frac{1}{n^2}\right)$
$$ \begin{aligned} & =\frac{1}{\lambda_0}\left(1-\frac{1}{n^2}\right) \\\\ \lambda & =\frac{\lambda_0}{1-\frac{1}{n^2}}=\frac{\lambda_0 n^2}{n^2-1} \\\\ & =\frac{\lambda_0\left(n^2-1\right)+1}{n^2-1} \\\\ & =\lambda_0\left(1+\frac{1}{n^2-1}\right) \end{aligned} $$
If $n^2-1=m^2$ ( $m$ is an integer)
then this is rarely true as $\left(n^2-1\right)$ cannot always be the square of an integer. Hence, option (C) is incorrect.
$$ \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{n^2}\right) n>2 $$
$\lambda$ is minimum for transition from $n_1 \rightarrow \infty$ to $n_2=2$.
$$ \frac{1}{\lambda_{\min }}=\frac{R}{4} $$
$\lambda$ is maximum for transition from $n_1=3$ to $n_2=2$
$$ \frac{1}{\lambda_{\max }}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5 \mathrm{R}}{36} $$
Dividing the above two equations, we get
$$ \frac{\lambda_{\max }}{\lambda_{\min }}=\frac{R}{4} \times \frac{36}{5 R}=\frac{9}{5} $$
Option (A) is correct.
For Lyman series,
$\lambda_{\min }=91.18 \mathrm{~nm}$
$\lambda_{\max }=121.57 \mathrm{~nm}$
For Balmer series,
$\lambda_{\min }=364.60 \mathrm{~nm}$
$\lambda_{\max }=656.30 \mathrm{~nm}$
For Pashen series,
$\lambda_{\min }=820.40 \mathrm{~nm}$
$\lambda_{\max }=1875 \mathrm{~nm}$
We can see from the above that there is no overlap between the wavelength ranges of Balmer and Paschen series. so, option (B) is incorrect.
Similarly wavelength ranges of Lyman and Balmer series also do not overlap. Hence, option (D) is correct.
Shortest wavelength of Lyman series
$$ \begin{aligned} \frac{1}{\lambda_0} & =R\left(1-\frac{1}{\infty}\right) \\\\ \lambda_0 & =\frac{1}{R} \end{aligned} $$
For Lyman series, $\frac{1}{\lambda}=\mathrm{R}\left(1-\frac{1}{n^2}\right)$
$$ \begin{aligned} & =\frac{1}{\lambda_0}\left(1-\frac{1}{n^2}\right) \\\\ \lambda & =\frac{\lambda_0}{1-\frac{1}{n^2}}=\frac{\lambda_0 n^2}{n^2-1} \\\\ & =\frac{\lambda_0\left(n^2-1\right)+1}{n^2-1} \\\\ & =\lambda_0\left(1+\frac{1}{n^2-1}\right) \end{aligned} $$
If $n^2-1=m^2$ ( $m$ is an integer)
then this is rarely true as $\left(n^2-1\right)$ cannot always be the square of an integer. Hence, option (C) is incorrect.
Comments (0)
