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JEE Main 202116 MarchMorning Shift
Exam
JEE Main
Year
2021
Date
16 March
Shift
Morning Shift
Subject
Physics
Chapter
Atomic Physics

JEE Main 2021 · 16 March · Morning Shift · Physics · Atomic Physics

The first three spectral lines of H -atom in the Balmer series are...

The first three spectral lines of HH-atom in the Balmer series are given λ1,λ2,λ3{\lambda}_{1}, {\lambda}_{2}, {\lambda}_{3} considering the Bohr atomic model, the wave lengths of first and third spectral lines (λ1λ3)\left(\dfrac{{\lambda}_{1}}{{\lambda}_{3}}\right) are related by a factor of approximately x×10−1x \times {10}^{- 1}. The value of xx, to the nearest integer, is ________.

Correct answer

0

Step-by-step solution

1λ1=Rz2(122−132)\dfrac{1}{{\lambda}_{1}} = R{z}^{2}\left(\dfrac{1}{{2}^{2}} - \dfrac{1}{{3}^{2}}\right)

1λ1=Rz2536\dfrac{1}{{\lambda}_{1}} = R{z}^{2}\dfrac{5}{36} ...(i). . . ( i )

For 3rd{3}^{rd} line

1λ3=Rz2(122−152)\dfrac{1}{{\lambda}_{3}} = R{z}^{2}\left(\dfrac{1}{{2}^{2}} - \dfrac{1}{{5}^{2}}\right)

1λ3=Rz221100\dfrac{1}{{\lambda}_{3}} = R{z}^{2}\dfrac{21}{100} ...(ii). . . ( ii )

(ii)+(i)( ii ) + ( i )

λ1λ3=21100×365=1.512=15.12×10−1\dfrac{{\lambda}_{1}}{{\lambda}_{3}} = \dfrac{21}{100} \times \dfrac{36}{5} = 1 . 512 = 15 . 12 \times {10}^{- 1}

x≈15x \approx 15

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