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(A) The frequency of a spectral line in a hydrogen atom is given by $F = R c \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right]$,where $R$ is the Rydber

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$F_1 - F_2$

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(A) The frequency of a spectral line in a hydrogen atom is given by $F = R c \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right]$,where $R$ is the Rydberg constant and $c$ is the speed of light.For the second line of the Lyman series $(n_f = 1, n_i = 3)$:$F_1 = R c \left[ \frac{1}{1^2} - \frac{1}{3^2} \right] = R c \left[ 1 - \frac{1}{9} \right] \quad \dots (i)$For the first line of the Balmer series $(n_f = 2, n_i = 3)$:$F_2 = R c \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] = R c \left[ \frac{1}{4} - \frac{1}{9} \right] \quad \dots (ii)$Subtracting equation $(ii)$ from equation $(i)$:$F_1 - F_2 = R c \left[ \left( 1 - \frac{1}{9} \right) - \left( \frac{1}{4} - \frac{1}{9} \right) \right]$$F_1 - F_2 = R c \left[ 1 - \frac{1}{4} \right]$The frequency of the first line of the Lyman series $(n_f = 1, n_i = 2)$ is:$F = R c \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = R c \left[ 1 - \frac{1}{4} \right]$Thus,the frequency of the first line of the Lyman series is $F = F_1 - F_2$.

Let $F_1$ be the frequency of the second line of the Lyman series and $F_2$ be the frequency of the first line of the Balmer series. Then,the frequency of the first line of the Lyman series is given by:

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