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For any given series of spectral lines of atomic hydrogen, let $\Delta \overline{v}=\overline{v}_{max}-\overline{v}_{min}$ be the difference in maximum and minimum frequencies in $cm^{-1}.$

The ratio $\Delta \overline{v}_{Lyman}/\Delta \overline{v}_{Balmer}$ is :

Option 1)
$4:1$

Option 2)
$9:4$

Option 3)
$5:4$

Option 4)
$27:5$

Answers (1)

S solutionqc

Lyman Series : $V_{max}\Rightarrow \infty \rightarrow 1$ (electron jump)

$V_{max}=Rc\left ( 1 \right )^{2}\left [ \frac{1}{12}-\frac{1}{\infty ^{2}} \right ]=Rc$

$V_{min}\Rightarrow 2\rightarrow 1$

$V_{min}=Rc\left ( 1 \right )^{2}\left [ \frac{1}{12}-\frac{1}{2 ^{2}} \right ]=\frac{3Rc}{4}$

$\left ( V_{max}-V_{min} \right )=Rc-\frac{3Rc }{4}$

$\left ( V_{max}-V_{min} \right )=\frac{Rc }{4}$

Balmen Series :

$V_{max}\Rightarrow \infty \rightarrow 2$

$V_{max}=Rc\left ( 1 \right )^{2}\left ( \frac{1}{2^{2}}-\frac{1}{\infty ^{2}} \right )=\frac{Rc}{4}$

$V_{min}\Rightarrow 3\rightarrow 2$

$V_{max}=Rc\left ( 1 \right )^{2}\left ( \frac{1}{2^{2}}-\frac{1}{3 ^{2}} \right )=\frac{9-4}{36}Rc=\frac{5Rc}{36}$

$V_{max}-V_{min}=\frac{Rc}{4}-\frac{5Rc}{36}=\frac{4Rc}{36}=\frac{Rc}{9}$

$Ratio=\frac{\Delta \overline{v}Lyman}{\Delta \overline{v}Balmer}=\frac{Rc}{4}\times \frac{9}{Rc}=9:4$

Option 1)

$4:1$

Option 2)

$9:4$

Option 3)

$5:4$

Option 4)

$27:5$

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