The wavelength of a spectral line emitted by hydrogen atom in the Balmer series is (16/3 R) ( R is Rydberg constant). What is the value of the principal quantum number of the state from which the transition takes place?

Question

The wavelength of a spectral line emitted by hydrogen atom in the Balmer series is (16/3 R) ( R is Rydberg constant). What is the value of the principal quantum number of the state from which the transition takes place? (A) 2 (B) 3 (C) 4 (D) 5

Tardigrade Admin · Accepted Answer

In spectral line series, wavelength is given by

\frac{1}{λ} = R (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}})

For Balmer transitions,

\Rightarrow \frac{1}{λ} = R (\frac{1}{4} - \frac{1}{n ^{2}})

Given,

λ = \frac{16}{3 R}

\Rightarrow \frac{3 R}{16} = R (\frac{1}{4} - \frac{1}{n ^{2}})

\Rightarrow \frac{3}{16} = \frac{1}{4} - \frac{1}{n ^{2}}

\Rightarrow \frac{1}{n ^{2}} = \frac{1}{4} - \frac{3}{16} = \frac{1}{16}

\Rightarrow n = 4

Q. The wavelength of a spectral line emitted by hydrogen atom in the Balmer series is 3R16​ ( R is Rydberg constant). What is the value of the principal quantum number of the state from which the transition takes place?

2

3

4

5

In spectral line series, wavelength is given by λ1​=R(nf2​1​−ni2​1​) For Balmer transitions, ⇒λ1​=R(41​−n21​) Given, λ=3R16​ ⇒163R​=R(41​−n21​) ⇒163​=41​−n21​ ⇒n21​=41​−163​=161​ ⇒n=4

Q. The wavelength of a spectral line emitted by hydrogen atom in the Balmer series is $\frac{16}{3 R}$ ( $R$ is Rydberg constant). What is the value of the principal quantum number of the state from which the transition takes place?