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Atoms and Nuclei question

2021 · 16 Mar · Shift 1 · Q70
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Atoms and Nuclei question

2021 · 16 Mar · Shift 1 · Q70

JEE MainPhysicsAtoms and NucleiNumerical+4 / −1
The first three spectral lines of H-atom in the Balmer series are given λ\lambdaλ 1, λ\lambdaλ 2, λ\lambdaλ 3 considering the Bohr atomic model, the wave lengths of first and third spectral lines (λ1λ3)\left( \frac{\lambda_{1} }{\lambda_{3} } \right)(λ3​λ1​​) are related by a factor of approximately 'x'×\times× 10 −-− 1. The value of x, to the nearest integer, is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 15

  1. Balmer series formula

For the Balmer series of hydrogen, the electron falls to n=2n=2n=2 from higher levels n=3,4,5,…n=3,4,5,\dotsn=3,4,5,…

Using 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​)

The first three Balmer lines are:

  • First line: n=3→2n=3 \to 2n=3→2 gives λ1\lambda_1λ1​
  • Second line: n=4→2n=4 \to 2n=4→2 gives λ2\lambda_2λ2​
  • Third line: n=5→2n=5 \to 2n=5→2 gives λ3\lambda_3λ3​
  1. Write expressions for λ1\lambda_1λ1​ and λ3\lambda_3λ3​

For λ1\lambda_1λ1​:

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,

λ1=365R\lambda_1=\frac{36}{5R}λ1​=5R36​

For λ3\lambda_3λ3​:

1λ3=R(14−125)=R(25−4100)=21R100\frac{1}{\lambda_3}=R\left(\frac{1}{4}-\frac{1}{25}\right) =R\left(\frac{25-4}{100}\right) =\frac{21R}{100}λ3​1​=R(41​−251​)=R(10025−4​)=10021R​

So,

λ3=10021R\lambda_3=\frac{100}{21R}λ3​=21R100​
  1. Find the ratio λ1λ3\dfrac{\lambda_1}{\lambda_3}λ3​λ1​​
λ1λ3=365R⋅21R100=36×21500=756500=1.512\frac{\lambda_1}{\lambda_3} =\frac{36}{5R}\cdot \frac{21R}{100} =\frac{36\times 21}{500} =\frac{756}{500} =1.512λ3​λ1​​=5R36​⋅10021R​=50036×21​=500756​=1.512
  1. Match with the given form

The question says the ratio is approximately of the form:

x×10−1x \times 10^{-1}x×10−1

Now,

1.512=15.12×10−11.512 = 15.12 \times 10^{-1}1.512=15.12×10−1

Hence,

x≈15x \approx 15x≈15

To the nearest integer,

15\boxed{15}15​
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