Question 98: Atoms & Nuclei

Question

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

Accepted Answer

For Balmer series
$$
\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{n^2}ight) n>2
$$
$\lambda$ is minimum for transition from $n_1 ightarrow \infty$ to $n_2=2$.
$$
\frac{1}{\lambda_{\min }}=\frac{R}{4}
$$
$\lambda$ is maximum for transition from $n_1=3$ to $n_2=2$
$$
\frac{1}{\lambda_{\max }}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}ight)=\frac{5 \mathrm{R}}{36}
$$
Dividing the above two equations, we get
$$
\frac{\lambda_{\max }}{\lambda_{\min }}=\frac{R}{4} 	imes \frac{36}{5 R}=\frac{9}{5}
$$
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
$$
\begin{aligned}
\frac{1}{\lambda_0} & =R\left(1-\frac{1}{\infty}ight) \\
\lambda_0 & =\frac{1}{R}
\end{aligned}
$$
For Lyman series, $\frac{1}{\lambda}=\mathrm{R}\left(1-\frac{1}{n^2}ight)$
$$
\begin{aligned}
& =\frac{1}{\lambda_0}\left(1-\frac{1}{n^2}ight) \\
\lambda & =\frac{\lambda_0}{1-\frac{1}{n^2}}=\frac{\lambda_0 n^2}{n^2-1} \\
& =\frac{\lambda_0\left(n^2-1ight)+1}{n^2-1} \\
& =\lambda_0\left(1+\frac{1}{n^2-1}ight)
\end{aligned}
$$
If $n^2-1=m^2$ ( $m$ is an integer)
then this is rarely true as $\left(n^2-1ight)$ cannot always be the square of an integer. Hence, option (C) is incorrect.

Answer

The ratio of the longest wavelength to the shortest wavelength in Balmer series is 9/5

Answer

There is an overlap between the wavelength ranges of Balmer and Paschen series

Answer

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

Answer

The wavelength ranges of Lyman and Balmer series do not overlap

Atoms & Nuclei

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

Choose the correct answer: