Atoms & Nuclei

JEE Main level, numerical answer, medium

137 of 239

Question 137 of 239

The first member of the Balmer series of hydrogen atom has a wavelength of 6561A∘6561A^{^{\circ}}. The wavelength of the second member of the Balmer series (in nm) is_____________

Show answer and solution

Answer: 486

1λ=RZ2(1n12−1n22)\dfrac{1}{\lambda }=R{Z}^{2}\left(\dfrac{1}{{n}_{1}^{2}}-\dfrac{1}{{n}_{2}^{2}}\right)
1λ1=R(1)2(122−142)=3R16\dfrac{1}{{\lambda }_{1}}=R{\left(1\right)}^{2}\left(\dfrac{1}{{2}^{2}}-\dfrac{1}{{4}^{2}}\right)=\dfrac{3R}{16}
λ2λ1=2027\dfrac{{\lambda }_{2}}{{\lambda }_{1}}=\dfrac{20}{27}
λ2=2027×6561A∘=4860A∘{\lambda }_{2}=\dfrac{20}{27}\times 6561A^{^{\circ}}=4860A^{^{\circ}}
=486nm=486nm

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.