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Atoms and Nuclei question

2020 · 8 Jan · Shift 2 · Q58
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Atoms and Nuclei question

2020 · 8 Jan · Shift 2 · Q58

JEE MainPhysicsAtoms and NucleiNumerical+4 / −1
The first member of the Balmer series of hydrogen atom has a wavelength of 6561 Å. The wavelength of the second member of the Balmer series (in nm) is:
Numerical answer
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Correct answer: 486

  1. Balmer series formula

For hydrogen, the Balmer series corresponds to transitions ending at n=2n=2n=2.

The wavelength is given by the Rydberg formula:

1λ=R(122−1n2)\frac{1}{\lambda} = R\left(\frac{1}{2^2} - \frac{1}{n^2}\right)λ1​=R(221​−n21​)
  • First member: transition n=3→2n=3 \to 2n=3→2
  • Second member: transition n=4→2n=4 \to 2n=4→2
  1. First member wavelength

Given:

λ1=6561 A˚\lambda_1 = 6561\,\text{\AA}λ1​=6561A˚

For the first member:

1λ1=R(14−19)=R(9−436)=5R36\frac{1}{\lambda_1} = R\left(\frac{1}{4} - \frac{1}{9}\right) = R\left(\frac{9-4}{36}\right) = \frac{5R}{36}λ1​1​=R(41​−91​)=R(369−4​)=365R​
  1. Second member wavelength

For the second member:

1λ2=R(14−116)=R(4−116)=3R16\frac{1}{\lambda_2} = R\left(\frac{1}{4} - \frac{1}{16}\right) = R\left(\frac{4-1}{16}\right) = \frac{3R}{16}λ2​1​=R(41​−161​)=R(164−1​)=163R​
  1. Take ratio

Divide the two expressions:

1λ21λ1=3R165R36=316⋅365=2720\frac{\frac{1}{\lambda_2}}{\frac{1}{\lambda_1}} = \frac{\frac{3R}{16}}{\frac{5R}{36}} = \frac{3}{16}\cdot\frac{36}{5} = \frac{27}{20}λ1​1​λ2​1​​=365R​163R​​=163​⋅536​=2027​

So,

λ1λ2=2720\frac{\lambda_1}{\lambda_2} = \frac{27}{20}λ2​λ1​​=2027​

Hence,

λ2=λ1⋅2027\lambda_2 = \lambda_1 \cdot \frac{20}{27}λ2​=λ1​⋅2720​

Substitute λ1=6561 A˚\lambda_1 = 6561\,\text{\AA}λ1​=6561A˚:

λ2=6561×2027\lambda_2 = 6561 \times \frac{20}{27}λ2​=6561×2720​

Since,

6561÷27=2436561 \div 27 = 2436561÷27=243

therefore,

λ2=243×20=4860 A˚\lambda_2 = 243 \times 20 = 4860\,\text{\AA}λ2​=243×20=4860A˚
  1. Convert to nm

Using

1 nm=10 A˚1\,\text{nm} = 10\,\text{\AA}1nm=10A˚

we get,

4860 A˚=486 nm4860\,\text{\AA} = 486\,\text{nm}4860A˚=486nm
  1. Final answer
486\boxed{486}486​
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