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Atoms and Nuclei question

2019 · 9 Apr · Shift 1 · Q49
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Atoms and Nuclei question

2019 · 9 Apr · Shift 1 · Q49

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be :
  1. A
    642.7 nm
  2. B
    488.9 nm
  3. C
    889.2 nm
  4. D
    388.9 nm
View written solutionFree

Correct answer: B

  1. Use the Rydberg formula for Balmer lines

For hydrogen,

1λ=R(122−1n2)\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)λ1​=R(221​−n21​)

for Balmer series, where the final state is n=2n=2n=2.

  1. First Balmer line: n=3→2n=3 \to 2n=3→2

Given wavelength:

λ1=660 nm\lambda_1=660\,\text{nm}λ1​=660nm

Now,

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​

So,

1660=5R36\frac{1}{660}=\frac{5R}{36}6601​=365R​
  1. Second Balmer line: n=4→2n=4 \to 2n=4→2

Let its wavelength be λ2\lambda_2λ2​. Then,

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 to eliminate RRR
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​

Hence,

1λ2=2720⋅1λ1\frac{1}{\lambda_2}=\frac{27}{20}\cdot \frac{1}{\lambda_1}λ2​1​=2027​⋅λ1​1​

So,

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

Substitute λ1=660 nm\lambda_1=660\,\text{nm}λ1​=660nm:

λ2=660×2027=1320027≈488.9 nm\lambda_2=660\times \frac{20}{27} =\frac{13200}{27} \approx 488.9\,\text{nm}λ2​=660×2720​=2713200​≈488.9nm
  1. Match with the options
λ2≈488.9 nm\lambda_2 \approx 488.9\,\text{nm}λ2​≈488.9nm

So the correct option is B.

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