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JEE Main 201910 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
10 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Structure of Atom

JEE Main 2019 · 10 April · Evening Shift · Chemistry · Structure of Atom

The ratio of the shortest wavelength of two spectral series of...

The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about 99. The spectral series are:

Options

  1. APaschen and Pfund
  2. BBalmer and Brackett
  3. CLyman and Paschen
  4. DBrackett and Pfund

Correct answer

C. Lyman and Paschen

Step-by-step solution

1λ2=RH(1n12−1n22)1λ1=RH(1m12−1m22)\dfrac{\dfrac{1}{{\lambda}_{2}} = {R}_{H}\left(1{n}_{1}^{2} - 1{n}_{2}^{2}\right)}{\dfrac{1}{{\lambda}_{1}} = {R}_{H}\left(1{m}_{1}^{2} - 1{m}_{2}^{2}\right)}

As for shortest wavelengths, n2andm2{n}_{2} and {m}_{2} is ∞\infty.
∴λ1λ2=91=m12n12∴ \dfrac{{\lambda}_{1}}{{\lambda}_{2}} = \dfrac{9}{1} = \dfrac{{m}_{1}^{2}}{{n}_{1}^{2}}

Now if m1=3andn1=1{m}_{1} = 3 and {n}_{1} = 1 it will justify the statement hence Lyman and Paschen (B)\left(B\right) is correct.
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