Ratio of longest wave lengths corresponding to Lyman and Balmer series in hydrogen spectrum is:

  • Option 1)

    \frac{9}{31}

  • Option 2)

    \frac{5}{27}

  • Option 3)

    \frac{3}{23}

  • Option 4)

    \frac{7}{29}

 

Answers (1)

As we learnt in

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\, \, \frac{1}{\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}{31}

Incorrect

Option 2)

\frac{5}{27}

Correct

Option 3)

\frac{3}{23}

Incorrect

Option 4)

\frac{7}{29}

Incorrect

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-
Buy Now
Exams
Articles
Questions