Atomic Structure Result Question 5-1
5. For any given series of spectral lines of atomic hydrogen, let $\Delta \bar{v}=\bar{v}{\text {max }}-\bar{v}{\text {min }}$ be the difference in maximum and minimum frequencies in $\mathrm{cm}^{-1}$. The ratio $\Delta \bar{v}{\text {Lyman }} / \Delta \bar{v}{\text {Balmer }}$ is
(2019 Main, 9 April I)
(a) $27: 5$
(b) $5: 4$
(c) $9: 4$
(d) $4: 1$
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Answer:
Correct Answer: 5. ( c )
Solution:
- For any given series of spectral lines of atomic hydrogen.
Let $\Delta \bar{v}=\bar{v}{\text {max }}-\bar{v}{\text {min }}$ be the difference in maximum and minimum frequencies in $\mathrm{cm}^{-1}$.
For Lyman series,
$\Delta \bar{v}=\bar{v}{\text {max }}-\bar{v}{\text {min }}$
General formula:
$\overline{\mathrm{v}}=109677\left[\dfrac{1}{n_i^2}-\dfrac{1}{n_f^2}\right]$
For Lyman $n_1=1, n_2=2,3, \ldots$
$\begin{aligned} & \bar{v}{\max }=109,677\left(\dfrac{1}{1}-\dfrac{1}{\infty}\right)=109,677\left(\dfrac{1}{1}-0\right)=109,677 \\ & \bar{v}{\min }=109,677\left(\dfrac{1}{1}-\dfrac{1}{(2)^2}\right) \\ & \Delta \bar{v}{\text {Lyman }}=\bar{v}{\max }-\bar{v}_{\min } \\ &=109,677-\left[\dfrac{109,677 \times 3}{4}\right]=\dfrac{109,677}{4} \end{aligned}$
For Balmer series,
$\overline{\mathrm{v}}_{\text {max }}=109,677\left(\dfrac{1}{(2)^2}-\dfrac{1}{\infty}\right) \Rightarrow \dfrac{109677}{4} $
$\overline{\mathrm{v}}_{\text {min }}=109,677\left(\dfrac{1}{(2)^2}-\dfrac{1}{(3)^2}\right) \Rightarrow \dfrac{109677 \times 5}{36}$
$\Delta \overline{\mathrm{v}}=\overline{\mathrm{v}}{\max }-\overline{\mathrm{v}}{\text {min }} $
$\Delta \overline{\mathrm{v}}_{\text {Balmer }}=\dfrac{109,677}{4}-\left[\dfrac{109,677}{36} \times 5\right]=109,677\left(\dfrac{1}{9}\right) $
$\dfrac{\Delta \overline{\mathrm{v}}{\text {Lyman }}}{\Delta \overline{\mathrm{v}}{\text {Balmer }}}=\dfrac{109,677 / 4}{109,677 / 9} $
$\dfrac{\Delta \overline{\mathrm{v}}{\text {Lyman }}}{\Delta \overline{\mathrm{v}}{\text {Balmer }}}=\dfrac{9}{4}$
$\therefore$ The ratio of $\dfrac{\Delta \bar{v}{\text {Lyman }}}{\Delta \bar{v}{\text {Balmer }}}$ is $9: 4$.
BETA
