What is the ratio of wave number of first line (lowest energy line) of Balmer series of H atomic spectrum to first line of its Brackett series?

Step 1:

For any line of the hydrogen spectrum, the wave number (reciprocal wavelength) is given by the Rydberg formula
$$\tilde{\nu}=R\left(\frac1{n_f^2}-\frac1{n_i^2}\right)$$
where
  • $$n_f$$ = quantum number of the lower (final) level,
  • $$n_i$$ = quantum number of the higher (initial) level,
  • $$R$$ = Rydberg constant.

Step 2:

Balmer series (first line)

Balmer series has $$n_f=2$$.
The first (lowest-energy) line corresponds to the transition $$n_i=3 \rightarrow n_f=2$$.

Therefore,
$$\tilde{\nu}_{\text{Balmer}} = R\left(\frac1{2^2}-\frac1{3^2}\right) =R\left(\frac14-\frac19\right) =R\left(\frac{9-4}{36}\right) =\frac{5R}{36}\,.$$

Brackett series (first line)

Brackett series has $$n_f=4$$.
Its first line comes from $$n_i=5 \rightarrow n_f=4$$.
Hence
$$\tilde{\nu}_{\text{Brackett}} = R\left(\frac1{4^2}-\frac1{5^2}\right) =R\left(\frac1{16}-\frac1{25}\right) =R\left(\frac{25-16}{400}\right) =\frac{9R}{400}\,.$$

Step 3:

Required ratio
$$\frac{\tilde{\nu}_{\text{Balmer}}}{\tilde{\nu}_{\text{Brackett}}} =\frac{\dfrac{5R}{36}}{\dfrac{9R}{400}} =\frac{5}{36}\times\frac{400}{9} =\frac{2000}{324} =\frac{500}{81}\approx6.17.$$

Express this ratio with the first term taken as $$5$$:

Let the ratio be written as $$5:x$$. 

Then, $$\frac{5}{x}=6.17 \;\Rightarrow\; x\approx\frac{5}{6.17}\approx0.81.$$

Thus the ratio of the wave numbers is $$5:0.81$$.

Option B $$\longrightarrow\ $$ $$5:0.81$$