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# Engineering

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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