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Atoms and Nuclei question

2024 · 6 Apr · Shift 2 · Q73
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Atoms and Nuclei question

2024 · 6 Apr · Shift 2 · Q73

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
The longest wavelength associated with Paschen series is : (Given RH=1.097×107SI\mathrm{R}_{\mathrm{H}}=1.097 \times 10^7 \mathrm{SI}RH​=1.097×107SI unit)
  1. A
    2.973×10−6 m2.973 \times 10^{-6} \mathrm{~m}2.973×10−6 m
  2. B
    1.876×10−6 m1.876 \times 10^{-6} \mathrm{~m}1.876×10−6 m
  3. C
    1.094×10−6 m1.094 \times 10^{-6} \mathrm{~m}1.094×10−6 m
  4. D
    3.646×10−6 m3.646 \times 10^{-6} \mathrm{~m}3.646×10−6 m
View written solutionFree

Correct answer: B

  1. Use the Rydberg formula for hydrogen spectrum

For a spectral line,

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

For the Paschen series, the final state is

n1=3n_1=3n1​=3

so

1λ=RH(132−1n22)\frac{1}{\lambda}=R_H\left(\frac{1}{3^2}-\frac{1}{n_2^2}\right)λ1​=RH​(321​−n22​1​)

with n2=4,5,6,…n_2=4,5,6,\dotsn2​=4,5,6,…

  1. Find which transition gives the longest wavelength

Since

λ∝1(19−1n22)\lambda \propto \frac{1}{\left(\frac{1}{9}-\frac{1}{n_2^2}\right)}λ∝(91​−n22​1​)1​

for the longest wavelength, the energy difference must be minimum.

That happens for the first line of the Paschen series:

n2=4→n1=3n_2=4 \to n_1=3n2​=4→n1​=3
  1. Substitute values
1λ=RH(19−116)\frac{1}{\lambda}=R_H\left(\frac{1}{9}-\frac{1}{16}\right)λ1​=RH​(91​−161​)

Compute the bracket:

19−116=16−9144=7144\frac{1}{9}-\frac{1}{16}=\frac{16-9}{144}=\frac{7}{144}91​−161​=14416−9​=1447​

So,

1λ=1.097×107×7144\frac{1}{\lambda}=1.097\times 10^7 \times \frac{7}{144}λ1​=1.097×107×1447​

Hence,

λ=1447×1.097×107\lambda=\frac{144}{7\times 1.097\times 10^7}λ=7×1.097×107144​
  1. Numerical calculation
7×1.097=7.6797\times 1.097 = 7.6797×1.097=7.679

Thus,

λ=1447.679×107\lambda = \frac{144}{7.679\times 10^7}λ=7.679×107144​ λ≈1.876×10−6 m\lambda \approx 1.876\times 10^{-6}\ \text{m}λ≈1.876×10−6 m
  1. Match with the options

This corresponds to:

Option B: 1.876×10−6 m1.876 \times 10^{-6}\,\text{m}1.876×10−6m

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