Question

The wave numbers of three spectral lines of hydrogen atom are considered. Identify the set of spectral lines belonging to the {Balmer series. (\(R\) = Rydberg constant)}

• A. \( \dfrac{5R}{36},\ \dfrac{8R}{9},\ \dfrac{15R}{16} \)
• B. \( \dfrac{7R}{144},\ \dfrac{3R}{16},\ \dfrac{16R}{255} \)
• C. \( \dfrac{3R}{4},\ \dfrac{3R}{16},\ \dfrac{7R}{144} \)
• D. \( \dfrac{5R}{36},\ \dfrac{3R}{16},\ \dfrac{21R}{100} \)

Hint:

For Balmer series, always factor out \(\dfrac{1}{4}\); if the remaining term is \(\dfrac{1}{n^2}\), it belongs to the series.

Answer 1

The Correct Option is D

Concept: The wave number of a hydrogen spectral line is given by the Rydberg formula: \[ \bar{\nu} = R\!\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right), \quad n_2>n_1 \] For the Balmer series: \[ n_1 = 2,\quad n_2 = 3,4,5,\ldots \] Thus, Balmer series wave numbers have the form: \[ \bar{\nu}=R\!\left(\frac{1}{4}-\frac{1}{n^2}\right) \]
Step 1: Check each expression 
\( \dfrac{5R}{36} \): \[ \frac{5}{36}=\frac{1}{4}-\frac{1}{9} \Rightarrow n_2=3 \quad \checkmark \] 
\( \dfrac{3R}{16} \): \[ \frac{3}{16}=\frac{1}{4}-\frac{1}{16} \Rightarrow n_2=4 \quad \checkmark \] 
\( \dfrac{21R}{100} \): \[ \frac{21}{100}=\frac{1}{4}-\frac{1}{25} \Rightarrow n_2=5 \quad \checkmark \] All three expressions correspond to transitions ending at \(n=2\). Final Answer: \[ \boxed{\text{Option (D)}} \]