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Atoms and Nuclei question

2022 · 26 Jul · Shift 1 · Q71
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Atoms and Nuclei question

2022 · 26 Jul · Shift 1 · Q71

JEE MainPhysicsAtoms and NucleiNumerical+4 / −1
In the hydrogen spectrum, λ\lambdaλ be the wavelength of first transition line of Lyman series. The wavelength difference will be "a λ\lambdaλ'' between the wavelength of 3rd 3^{\text {rd }}3rd  transition line of Paschen series and that of 2nd 2^{\text {nd }}2nd  transition line of Balmer series where a=‾\mathrm{a}=\underline{\hspace{2cm}}a=​.
Numerical answer
View written solutionFree

Correct answer: 5

  1. Use the Rydberg formula

For hydrogen spectrum,

1λ=R(1n12−1n22),n2>n1\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right), \qquad n_2>n_1λ1​=R(n12​1​−n22​1​),n2​>n1​

where RRR is the Rydberg constant.

  1. Find the wavelength λ\lambdaλ of the first line of Lyman series

Lyman series means transition to n1=1n_1=1n1​=1.

First line corresponds to n2=2→n1=1n_2=2 \to n_1=1n2​=2→n1​=1. So,

1λ=R(1−122)=R(1−14)=3R4\frac{1}{\lambda}=R\left(1-\frac{1}{2^2}\right)=R\left(1-\frac14\right)=\frac{3R}{4}λ1​=R(1−221​)=R(1−41​)=43R​

Hence,

λ=43R\lambda=\frac{4}{3R}λ=3R4​
  1. Find wavelength of 3rd transition line of Paschen series

Paschen series means transition to n1=3n_1=3n1​=3.

The lines are:

  • 1st line: 4→34 \to 34→3
  • 2nd line: 5→35 \to 35→3
  • 3rd line: 6→36 \to 36→3

Thus,

1λP=R(132−162)=R(19−136)=R(4−136)=R12\frac{1}{\lambda_P}=R\left(\frac{1}{3^2}-\frac{1}{6^2}\right) =R\left(\frac19-\frac{1}{36}\right) =R\left(\frac{4-1}{36}\right) =\frac{R}{12}λP​1​=R(321​−621​)=R(91​−361​)=R(364−1​)=12R​

So,

λP=12R\lambda_P=\frac{12}{R}λP​=R12​
  1. Find wavelength of 2nd transition line of Balmer series

Balmer series means transition to n1=2n_1=2n1​=2.

The lines are:

  • 1st line: 3→23 \to 23→2
  • 2nd line: 4→24 \to 24→2

Thus,

1λB=R(122−142)=R(14−116)=R(4−116)=3R16\frac{1}{\lambda_B}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right) =R\left(\frac14-\frac{1}{16}\right) =R\left(\frac{4-1}{16}\right) =\frac{3R}{16}λB​1​=R(221​−421​)=R(41​−161​)=R(164−1​)=163R​

So,

λB=163R\lambda_B=\frac{16}{3R}λB​=3R16​
  1. Find the wavelength difference

Required difference:

λP−λB=12R−163R\lambda_P-\lambda_B=\frac{12}{R}-\frac{16}{3R}λP​−λB​=R12​−3R16​

Taking LCM,

λP−λB=36−163R=203R\lambda_P-\lambda_B=\frac{36-16}{3R}=\frac{20}{3R}λP​−λB​=3R36−16​=3R20​

Now compare with

λ=43R\lambda=\frac{4}{3R}λ=3R4​

So,

203R=aλ=a(43R)\frac{20}{3R}=a\lambda=a\left(\frac{4}{3R}\right)3R20​=aλ=a(3R4​)

Hence,

a=204=5a=\frac{20}{4}=5a=420​=5
  1. Final answer
a=5\boxed{a=5}a=5​

The derived answer matches the stored correct answer.

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