Question

For any given series of spectral lines of atomic hydrogen, let $\overrightarrow{\Delta V}={{\overrightarrow{V}}_{\max }}-{{\overrightarrow{V}}_{\min }}$ be the difference in maximum and minimum frequencies in $c{{m}^{-1}}$. The ratio of ${{\overrightarrow{V}}_{Lyman}}/{{\overrightarrow{V}}_{Balmer}}$ is:
A. $27:5$
B. $4:1$
C. $5:4$
D. $9:4$

Answer 1

Hint: To solve this question, you should know that for the Lyman series, the value of the first spectral line starts from one whereas for the Balmer series, the value of the spectral line starts from two. By using the formula regarding the frequency for both the series, you can compare and find out the ratio.

Complete step by step solution:
Given that, For any given series of spectral lines of atomic hydrogen, let $\overrightarrow{\Delta V}={{\overrightarrow{V}}_{\max }}-{{\overrightarrow{V}}_{\min }}$ be the difference in maximum and minimum frequencies in .
The formula for frequency can also be written as:
$V=CR\left( \dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{n_{2}^{2}} \right)$
Where, V is the frequency, C is constant and R is Rydberg constant, ${{n}_{1}}$ and ${{n}_{2}}$ are the spectral line number. For the Lyman series, we should know that the first spectral line equals one and the second spectral line will continue as two, three, four and so on (for minimum frequency), while it will be infinite for the maximum frequency. Therefore, the maximum frequency will be ${{V}_{\max }}=CR\left( \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{\infty }^{2}}} \right)$. And for the minimum frequency it will be ${{V}_{\min }}=CR\left( \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{2}^{2}}} \right)$. So, the change in frequency for Lyman series can be represented as:
$\Delta {{V}_{Lyman}}=CR\left( \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{\infty }^{2}}} \right)-CR\left( \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{2}^{2}}} \right)=CR\left( 1-\dfrac{3}{4} \right)=CR\left( \dfrac{1}{4} \right)$
Whereas, for the Balmer series, the first spectral line will start from one and the second spectral line will start from three, four, and so on. So, the maximum frequency in case of Balmer series will be ${{V}_{Max}}=\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{\infty }^{2}}} \right)$
And the minimum frequency will be ${{V}_{Min}}=\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{3}^{2}}} \right)$. So, the change in frequency for Balmer series will be:
$\Delta {{V}_{Balmer}}=CR\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{\infty }^{2}}} \right)-CR\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{3}^{2}}} \right)=\dfrac{CR}{4}\left( 1-\dfrac{5}{9} \right)=\dfrac{CR}{9}$
Now, when we take the ratio of frequency of Lyman and Balmer, we get:
$\dfrac{{{V}_{Lyman}}}{{{V}_{Balmer}}}=\dfrac{CR}{4}\times \dfrac{9}{CR}=\dfrac{9}{4}=9:4$
So, the ratio of ${{V}_{Lyman}}/{{V}_{Balmer}}$ is $9:4$.

Hence, the correct option is D.

Note: It is important to note that the Balmer series is a series of spectral lines in the electromagnetic spectrum which lies in the visible region of the spectrum. Transition from first shell to any other shell is referred to as Lyman series while transition from second shell to any other shell is referred to as the Balmer series.