Question

If $\lambda$ is the longest wavelength of Lyman series, the shortest wavelength line that of the Balmer series is:
A. $5\lambda$
B. $2\lambda$
C. $4\lambda$
D. $3\lambda$

Answer 1

Hint: In this question you are asked to find the shortest wavelength line of the Balmer series, which can be find by putting the values of the values of n1 and n2 for the shortest wavelength in the Balmer series. It is important that you should know the general formula for wavelength to find the shortest wavelength of the Balmer series as well as longest wavelength of the Lyman series.

Formula used:
For longest wavelength in Lyman series (i.e., first line), the energy difference in two states showing transition should be minimum, i.e, $n_2$ = 2
${\displaystyle \frac{1}{\lambda }}=R[{\displaystyle \frac{1}{n_1^2}}-{\displaystyle \frac{1}{n_2^2}}]$
Where, $R$ is the Rydberg constant.
$\lambda$ is the wavelength.

Complete step by step answer:
General formula for wavelength of Lyman series is given as,
${\displaystyle \frac{1}{\lambda }}=R[{\displaystyle \frac{1}{n_1^2}}-{\displaystyle \frac{1}{n_2^2}}]$
Now, formula for longest wavelength of Lyman series is given as,
$n_1$ = 1 and $n_1$ = 2
$$\begin{gathered} {\displaystyle \frac{1}{{\lambda }_L}}=R[{\displaystyle \frac{1}{1^2}}-{\displaystyle \frac{1}{2^2}}] \\ \Rightarrow {\displaystyle \frac{1}{{\lambda }_L}}=R[{\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{4}}] \\ \Rightarrow {\displaystyle \frac{1}{{\lambda }_L}}={\displaystyle \frac{3R}{4}} \end{gathered}$$
Now according to the question, If $\lambda$is the longest wavelength of Lyman series
therefore,
$$\begin{gathered} {\lambda }_L={\displaystyle \frac{4}{3R}}=\lambda \\ \Rightarrow {\displaystyle \frac{4}{R}}=3\lambda ...(i) \end{gathered}$$
Now, formula for the shortest wavelength line that of the Balmer series is given as,
$n_1$ =2 and $n_2=\mathrm{\infty }$
$$\begin{gathered} {\displaystyle \frac{1}{{\lambda }_B}}=R[{\displaystyle \frac{1}{2^2}}-{\displaystyle \frac{1}{{\mathrm{\infty }}^2}}] \\ \Rightarrow {\displaystyle \frac{1}{{\lambda }_B}}={\displaystyle \frac{R}{4}} \\ \Rightarrow {\lambda }_B={\displaystyle \frac{4}{R}}...(ii) \end{gathered}$$
Now, from equations (i) and (ii), we get,
$\therefore {\lambda }_B=3\lambda$

Hence, option D is the correct option.

Note:It is important to note that the Lyman series is the first series while the Balmer series is the second series, in a hydrogen atom. Most of the students tend to make mistakes in the lowest states of these two states. The lowest state in the Lyman series is 1 while in the Balmer series it is 2. Again, it should be remembered that the spectral lines of Lyman series are ultra-violet while the spectral lines of Balmer series are visible.