Atoms & Nuclei

JEE Main level, single correct, medium

98 of 239

Question 98 of 239

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

Show answer and solution

Answer: (B) 1217

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}
λmax\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}
λmax\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

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Atoms: Rutherford, Bohr and X-rays simulationDrag the sliders and watch the numbers in this question change.Revise Modern Physics firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Atoms & Nuclei (Atomic Spectrum), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.