Taking the wavelength of first Balmer line in hydrogen spectrum ( $\mathrm{n}=3$ to $\mathrm{n}=2$ ) as 660 nm , the wavelength of the $2^{\text {nd }}$ Balmer line ( $\mathrm{n}=4$ to $\mathrm{n}=2$ ) will be

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Modern Physics

Taking the wavelength of first Balmer line in hydrogen spectrum (

n = 3

n = 2

) as 660 nm , the wavelength of the

2^{nd}

Balmer line (

n = 4

n = 2

) will be

(A)889.2 nm
(B)642.7 nm
(C)488.9 nm✓ Correct
(D)388.9 nm

Answer: (C)

Solution

For the first Balmer line

\frac{1}{660} = R (\frac{1}{4} - \frac{1}{9}) = \frac{5 R}{36}

, and for the second Balmer line

\frac{1}{λ} = R (\frac{1}{4} - \frac{1}{16}) = \frac{3 R}{16}

. Dividing the two relations gives

λ = 660 \times \frac{20}{27} \approx 488.9 nm

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option

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