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JEE Main 202622 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Evening Shift
Subject
Physics
Chapter
Atomic Physics

JEE Main 2026 · 22 January · Evening Shift · Physics · Atomic Physics

The smallest wavelength of Lyman series is 91 nm. The difference...

The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly ____\_\_\_\_ nm .

Options

  1. A1550
  2. B1217
  3. C1875
  4. D1784

Correct answer

B. 1217

Step-by-step solution

Smallest wavelength of lyman
1λ=R(112−1∞2)\dfrac{1}{\lambda} = R\left(\dfrac{1}{1^2} - \dfrac{1}{\infty^2}\right)
R=1λ=191 nm−1R = \dfrac{1}{\lambda} = \dfrac{1}{91} \text{ nm}^{-1}
\lambda_\max for balmer series
n1=2→n2=3n_1 = 2 \rightarrow n_2 = 3
1λB=R(14−19)\dfrac{1}{\lambda_B} = R\left(\dfrac{1}{4} - \dfrac{1}{9}\right)
1λB=191(536)\dfrac{1}{\lambda_B} = \dfrac{1}{91}\left(\dfrac{5}{36}\right)
λB=(91×365)=655.2 nm\lambda_B = \left(\dfrac{91 \times 36}{5}\right) = 655.2 \text{ nm}
\lambda_\max paschen
n1=3→n2=4n_1 = 3 \rightarrow n_2 = 4
1λp=191(132−142)=191×7144\dfrac{1}{\lambda_p} = \dfrac{1}{91}\left(\dfrac{1}{3^2} - \dfrac{1}{4^2}\right) = \dfrac{1}{91} \times \dfrac{7}{144}
λp=(91×1447)=1872 nm\lambda_p = \left(\dfrac{91 \times 144}{7}\right) = 1872 \text{ nm}
Δλ=λP−λB=1872−655.2\Delta\lambda = \lambda_P - \lambda_B = 1872 - 655.2
Δλ=1216.8\Delta\lambda = 1216.8
Δλ≈1217\Delta\lambda \approx 1217
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