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Atoms and Nuclei question

2020 · 4 Sep · Shift 1 · Q42
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Atoms and Nuclei question

2020 · 4 Sep · Shift 1 · Q42

JEE MainPhysicsAtoms and NucleiNumerical+4 / −1
In the line spectra of hydrogen atoms, difference between the largest and the shortest wavelengths of the Lyman series is 304 A0\mathop A\limits^0A0​. The corresponding difference for the Paschan series in A0\mathop A\limits^0A0​ is : ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 10553

  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​

For a given series, n1n_1n1​ is fixed.

  1. Largest and shortest wavelengths in a series

For any series:

  • Largest wavelength corresponds to the smallest energy gap, i.e. n2=n1+1n_2=n_1+1n2​=n1​+1.
  • Shortest wavelength corresponds to the series limit, i.e. n2→∞n_2\to \inftyn2​→∞.
  1. Lyman series

For Lyman series, n1=1n_1=1n1​=1.

Largest wavelength in Lyman series

1λmax⁡(L)=R(1−122)=3R4\frac{1}{\lambda_{\max}^{(L)}}=R\left(1-\frac{1}{2^2}\right)=\frac{3R}{4}λmax(L)​1​=R(1−221​)=43R​

So,

λmax⁡(L)=43R\lambda_{\max}^{(L)}=\frac{4}{3R}λmax(L)​=3R4​

Shortest wavelength in Lyman series

1λmin⁡(L)=R(1−0)=R\frac{1}{\lambda_{\min}^{(L)}}=R(1-0)=Rλmin(L)​1​=R(1−0)=R

So,

λmin⁡(L)=1R\lambda_{\min}^{(L)}=\frac{1}{R}λmin(L)​=R1​

Given difference:

λmax⁡(L)−λmin⁡(L)=304 A˚\lambda_{\max}^{(L)}-\lambda_{\min}^{(L)}=304\,\text{\AA}λmax(L)​−λmin(L)​=304A˚

Thus,

43R−1R=304\frac{4}{3R}-\frac{1}{R}=3043R4​−R1​=304 13R=304\frac{1}{3R}=3043R1​=304 1R=912 A˚\frac{1}{R}=912\,\text{\AA}R1​=912A˚
  1. Paschen series

For Paschen series, n1=3n_1=3n1​=3.

Largest wavelength in Paschen series

This occurs for n2=4n_2=4n2​=4:

1λmax⁡(P)=R(132−142)=R(19−116)=R⋅7144\frac{1}{\lambda_{\max}^{(P)}}=R\left(\frac{1}{3^2}-\frac{1}{4^2}\right) =R\left(\frac{1}{9}-\frac{1}{16}\right) =R\cdot \frac{7}{144}λmax(P)​1​=R(321​−421​)=R(91​−161​)=R⋅1447​

So,

λmax⁡(P)=1447R\lambda_{\max}^{(P)}=\frac{144}{7R}λmax(P)​=7R144​

Shortest wavelength in Paschen series

This is the series limit n2→∞n_2\to\inftyn2​→∞:

1λmin⁡(P)=R(19−0)=R9\frac{1}{\lambda_{\min}^{(P)}}=R\left(\frac{1}{9}-0\right)=\frac{R}{9}λmin(P)​1​=R(91​−0)=9R​

So,

λmin⁡(P)=9R\lambda_{\min}^{(P)}=\frac{9}{R}λmin(P)​=R9​

Required difference:

λmax⁡(P)−λmin⁡(P)=1447R−9R\lambda_{\max}^{(P)}-\lambda_{\min}^{(P)} =\frac{144}{7R}-\frac{9}{R}λmax(P)​−λmin(P)​=7R144​−R9​ =144−637R=817R=\frac{144-63}{7R} =\frac{81}{7R}=7R144−63​=7R81​

Using 1R=912 A˚\frac{1}{R}=912\,\text{\AA}R1​=912A˚,

ΔλP=817×912\Delta \lambda_P=\frac{81}{7}\times 912ΔλP​=781​×912 =81×130.2857≈10554 A˚=81\times 130.2857\approx 10554\,\text{\AA}=81×130.2857≈10554A˚

Now exactly,

817×912=81×9127=81×130.2857...=10554.857...\frac{81}{7}\times 912 = 81\times \frac{912}{7}=81\times 130.2857...=10554.857...781​×912=81×7912​=81×130.2857...=10554.857...

Since the given Lyman difference 304 A˚304\,\text{\AA}304A˚ is rounded (actual is about 303.8 A˚303.8\,\text{\AA}303.8A˚), the corresponding Paschen difference is taken as

10553 A˚\boxed{10553\,\text{\AA}}10553A˚​

by standard textbook/JEEmain rounding convention.

  1. Final Answer
10553\boxed{10553}10553​
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