If 917Å be the lowest… - JEE Main 2023 Physics PYQ

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.

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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,

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

Did you get this right?

Solution

Your note

JEE Main Modern Physics in other years

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Solution

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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