The wave number of first emission line in the atomic spectrum of hydrogen in the Balmer series is

Question

The wave number of first emission line in the atomic spectrum of hydrogen in the Balmer series is (A) (5 R/36)cm- 1 (B) (3 R/4)cm- 1 (C) (7 R/144)cm- 1 (D) (9 R/400)cm- 1

Tardigrade Admin · Accepted Answer

The Balmer Series of spectral lines occurs when electrons transition from an energy level higher than

n = 3

back down to

n = 2

.
Balmer series starts from

n = 2

. So first emission line in atomic spectrum of

H

is

2 \to 3

\frac{1}{λ} = R Z^{2} [\frac{1}{n _{1}^{2}} \frac{- 1}{n _{2}^{2}}]

\overset{ν}{ˉ} = R [\frac{1}{2 ^{2}} - \frac{1}{3 ^{2}}] c m^{- 1} = \frac{5 R}{36} c m^{- 1}

Q. The wave number of first emission line in the atomic spectrum of hydrogen in the Balmer series is

$\frac{5 R}{36} c m^{- 1}$

$\frac{3 R}{4} c m^{- 1}$

$\frac{7 R}{144} c m^{- 1}$

$\frac{9 R}{400} c m^{- 1}$

Q. The wave number of first emission line in the atomic spectrum of hydrogen in the Balmer series is

365R​cm−1

43R​cm−1

1447R​cm−1

4009R​cm−1

The Balmer Series of spectral lines occurs when electrons transition from an energy level higher than n=3 back down to n=2. Balmer series starts from n=2. So first emission line in atomic spectrum of H is 2→3 λ1​=RZ2[n12​1​n22​−1​] νˉ=R[221​−321​]cm−1=365R​cm−1

$\frac{5 R}{36} c m^{- 1}$

$\frac{3 R}{4} c m^{- 1}$

$\frac{7 R}{144} c m^{- 1}$

$\frac{9 R}{400} c m^{- 1}$