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  • Try this! In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is :

In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is :

  • Option 1)

    \frac{9} {\text{4}}

  • Option 2)

    \frac{{27}} {\text{5}}

  • Option 3)

    \frac{5} {{\text{27}}}

  • Option 4)

    \frac{4} {\text{9}}

 

Answers (1)

best_answer

 

Energy emitted due to transition of electron -

Delta E= Rhcz^{2}left ( frac{1}{n_{f}, ^{2}}-frac{1}{n_{i}, ^{2}} ight )

frac{1}{lambda }= Rz^{2}left ( frac{-1}{n_{i}, ^{2}}+frac{1}{n_{f}, ^{2}} ight )

- wherein

R= R hydberg: constant

n_{i}= initial state \n_{f}= final : state

 

 Longest wavelength in balmer series is corresponding to transition between 

Lets say it is \lambda _{1}

3\rightarrow 2 \: Then\:\lambda _{1}=R\left ( \frac{1}{4}-\frac{1}{9} \right )=\frac{5}{36}R

Longest wavelength in lymen series is corresponding to transition between 2\rightarrow 1

Lets say it is \lambda _{2}

Then \frac{1}{\lambda _{2}}-R\left ( 1-\frac{1}{4} \right )=\frac{3R}{4}

\frac{\lambda _{lymen}}{\lambda _{Balmer}}=\frac{\frac{4}{3R}}{\frac{36}{5R}} = \frac{4}{3R}\times \frac{5R}{36}=\frac{5}{27}

 


Option 1)

\frac{9} {\text{4}}

This is incorrect option

Option 2)

\frac{{27}} {\text{5}}

This is incorrect option

Option 3)

\frac{5} {{\text{27}}}

This is correct option

Option 4)

\frac{4} {\text{9}}

This is incorrect option

Posted by

prateek

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