JEE Main 2023 — Physics Modern Physics

2023integereasy

If $917Å$ be the lowest wavelength of Lyman series then the lowest wavelength of Balmer series will be $Å$ .

Official previous-year question

Held on 10 Apr 2023 · Verified 6 Jul 2026.

Did you get this right?

Solution

Using Rydberg's formula,

$\frac{1}{\lambda }=R{Z}^{2}(\frac{1}{{{n}^{2}}_{1}}-\frac{1}{{{n}^{2}}_{2}})$

For the lowest wavelength in Lyman series, ${n}_{1}=1,{n}_{2}=\infty$ .

$\frac{1}{\lambda }=R{Z}^{2}(\frac{1}{{1}^{2}}-\frac{1}{{\infty }^{2}})=R{Z}^{2}$

For Balmer series, ${n}_{1}=2,{n}_{2}=\infty$

$\frac{1}{\lambda '}=R{Z}^{2}(\frac{1}{{2}^{2}}-\frac{1}{{\infty }^{2}})=\frac{R{Z}^{2}}{4}$

So,

$\frac{1}{\lambda '}=\frac{1}{4\lambda } \Rightarrow \lambda '=4\times 917\overset{}{Å=3668Å}$

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JEE Main 2023 — Physics Modern Physics

2023integereasy

If $917Å$ be the lowest wavelength of Lyman series then the lowest wavelength of Balmer series will be $Å$ .

Official previous-year question

Held on 10 Apr 2023 · Verified 6 Jul 2026.

Did you get this right?

Solution

Using Rydberg's formula,

$\frac{1}{\lambda }=R{Z}^{2}(\frac{1}{{{n}^{2}}_{1}}-\frac{1}{{{n}^{2}}_{2}})$

For the lowest wavelength in Lyman series, ${n}_{1}=1,{n}_{2}=\infty$ .

$\frac{1}{\lambda }=R{Z}^{2}(\frac{1}{{1}^{2}}-\frac{1}{{\infty }^{2}})=R{Z}^{2}$

For Balmer series, ${n}_{1}=2,{n}_{2}=\infty$

$\frac{1}{\lambda '}=R{Z}^{2}(\frac{1}{{2}^{2}}-\frac{1}{{\infty }^{2}})=\frac{R{Z}^{2}}{4}$

So,

$\frac{1}{\lambda '}=\frac{1}{4\lambda } \Rightarrow \lambda '=4\times 917\overset{}{Å=3668Å}$

JEE Main 2023 — Physics Modern Physics

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Solution

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JEE Main Modern Physics in other years

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JEE Main 2023 — Physics Modern Physics

Did you get this right?

Solution

Your note

JEE Main Modern Physics in other years

More JEE Main Modern Physics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Modern Physics