JEE Main 2026 — Chemistry Physical Chemistry

The wave number $\bar{\nu}$νˉ of a spectral line in the hydrogen emission spectrum is given by the Rydberg formula:

$$\bar{\nu} = R_H \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)$$νˉ=RH​(n12​1​−n22​1​)

For the first line (lowest energy) of the Balmer series, the transition is from $n_2 = 3$n2​=3 to $n_1 = 2$n1​=2:

$$\bar{\nu}_{Balmer} = R_H \left( \dfrac{1}{2^2} - \dfrac{1}{3^2} \right) = R_H \left( \dfrac{1}{4} - \dfrac{1}{9} \right) = \dfrac{5R_H}{36}$$νˉBalmer​=RH​(221​−321​)=RH​(41​−91​)=365RH​​

For the first line of the Brackett series, the transition is from $n_2 = 5$n2​=5 to $n_1 = 4$n1​=4:

$$\bar{\nu}_{Brackett} = R_H \left( \dfrac{1}{4^2} - \dfrac{1}{5^2} \right) = R_H \left( \dfrac{1}{16} - \dfrac{1}{25} \right) = \dfrac{9R_H}{400}$$νˉBrackett​=RH​(421​−521​)=RH​(161​−251​)=4009RH​​

Taking the ratio of the two wave numbers:

$$\dfrac{\bar{\nu}_{Balmer}}{\bar{\nu}_{Brackett}} = \dfrac{\dfrac{5R_H}{36}}{\dfrac{9R_H}{400}} = \dfrac{5}{36} \times \dfrac{400}{9} = \dfrac{500}{81}$$νˉBrackett​νˉBalmer​​=4009RH​​365RH​​​=365​×9400​=81500​

To express this ratio in the form $5 : x$5:x, we divide the numerator and the denominator by $100$100:

$$\dfrac{500}{81} = \dfrac{5}{0.81}$$81500​=0.815​

The ratio is $5 : 0.81$5:0.81.

Answer: $5:0.81$5:0.81