If the first line of Lyman series has a wavel | vedclass

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(C) From the hydrogen spectrum, when an electron transitions from $n_2$ orbit to $n_1$ orbit, the emitted wavelength $\lambda$ is given by the Rydberg

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$6563$

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(C) From the hydrogen spectrum, when an electron transitions from $n_2$ orbit to $n_1$ orbit, the emitted wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Lyman series, $n_1 = 1$. The first line corresponds to $n_2 = 2$. Thus, $\frac{1}{\lambda_L} = R Z^2 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R Z^2 \left( 1 - \frac{1}{4} ight) = \frac{3}{4} R Z^2$.For the Balmer series, $n_1 = 2$. The first line corresponds to $n_2 = 3$. Thus, $\frac{1}{\lambda_B} = R Z^2 \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R Z^2 \left( \frac{1}{4} - \frac{1}{9} ight) = R Z^2 \left( \frac{5}{36} ight)$.Dividing the expression for $\frac{1}{\lambda_L}$ by $\frac{1}{\lambda_B}$, we get: $\frac{\lambda_B}{\lambda_L} = \frac{3/4}{5/36} = \frac{3}{4} 	imes \frac{36}{5} = \frac{27}{5} = 5.4$.Given $\lambda_L = 1215.4 	ext{ Å}$, we have $\lambda_B = 5.4 	imes 1215.4 	ext{ Å} = 6563.16 	ext{ Å} \approx 6563 	ext{ Å}$.

If the first line of Lyman series has a wavelength $1215.4 \text{ Å}$, the first line of Balmer series is approximately (in $\text{ Å}$)

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