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Atoms and Nuclei question

2011 · Shift 1 · Q58
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Atoms and Nuclei question

2011 · Shift 1 · Q58

JEE AdvancedPhysicsAtoms and NucleiMCQ+3 / −1
The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Ao\mathop A\limits^oAo​. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is
  1. A
    1215 Ao\mathop A\limits^oAo​
  2. B
    1640 Ao\mathop A\limits^oAo​
  3. C
    2430 Ao\mathop A\limits^oAo​
  4. D
    4687 Ao\mathop A\limits^oAo​
View written solutionFree

Correct answer: A

  1. Use the Rydberg formula for hydrogen-like atoms:
1λ=RZ2(1n12−1n22)\frac{1}{\lambda}=R Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)λ1​=RZ2(n12​1​−n22​1​)

where:

  • ZZZ = atomic number,
  • n2>n1n_2 > n_1n2​>n1​,
  • Balmer series means transitions to n1=2n_1=2n1​=2.
  1. First spectral line of Balmer series for hydrogen:

For hydrogen, Z=1Z=1Z=1.

First Balmer line corresponds to transition:

n2=3→n1=2n_2=3 \to n_1=2n2​=3→n1​=2

So,

1λH=R(122−132)=R(14−19)=R(536)\frac{1}{\lambda_H}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) =R\left(\frac{1}{4}-\frac{1}{9}\right) =R\left(\frac{5}{36}\right)λH​1​=R(221​−321​)=R(41​−91​)=R(365​)

Given:

λH=6561 A˚\lambda_H=6561\,\text{\AA}λH​=6561A˚
  1. Second spectral line of Balmer series for singly ionized helium:

Singly ionized helium is He+\mathrm{He}^+He+, which is hydrogen-like with:

Z=2Z=2Z=2

Second Balmer line corresponds to transition:

n2=4→n1=2n_2=4 \to n_1=2n2​=4→n1​=2

Thus,

1λHe+=R(2)2(122−142)=4R(14−116)=4R(316)=3R4\frac{1}{\lambda_{He^+}}=R(2)^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right) =4R\left(\frac{1}{4}-\frac{1}{16}\right) =4R\left(\frac{3}{16}\right) =\frac{3R}{4}λHe+​1​=R(2)2(221​−421​)=4R(41​−161​)=4R(163​)=43R​
  1. Take ratio with the hydrogen Balmer line:

For hydrogen first Balmer line:

1λH=5R36\frac{1}{\lambda_H}=\frac{5R}{36}λH​1​=365R​

For He+\mathrm{He}^+He+ second Balmer line:

1λHe+=3R4\frac{1}{\lambda_{He^+}}=\frac{3R}{4}λHe+​1​=43R​

Therefore,

λHe+λH=5R363R4=536⋅43=527\frac{\lambda_{He^+}}{\lambda_H} =\frac{\frac{5R}{36}}{\frac{3R}{4}} =\frac{5}{36}\cdot\frac{4}{3} =\frac{5}{27}λH​λHe+​​=43R​365R​​=365​⋅34​=275​

So,

λHe+=6561×527\lambda_{He^+}=6561\times \frac{5}{27}λHe+​=6561×275​

Now,

6561÷27=2436561\div 27=2436561÷27=243

Hence,

λHe+=243×5=1215 A˚\lambda_{He^+}=243\times 5=1215\,\text{\AA}λHe+​=243×5=1215A˚
  1. Match with options:
1215 A˚\boxed{1215\,\text{\AA}}1215A˚​

So the correct option is A.

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