The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength of t

Question

The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is:

Accepted Answer

The wavelength ($\lambda$) of a spectral line is given by the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}ight)$.

1.  **Hydrogen atom (Z=1):** The first line of the Balmer series corresponds to the transition from $n_2=3$ to $n_1=2$. 
    $\frac{1}{\lambda_H} = R(1)^2 \left(\frac{1}{2^2} - \frac{1}{3^2}ight) = R\left(\frac{1}{4} - \frac{1}{9}ight) = \frac{5R}{36}$.
    Given $\lambda_H = 6561 \mathring{A}$.

2.  **Singly ionized Helium atom (He$^+$, Z=2):** The second line of the Balmer series corresponds to the transition from $n_2=4$ to $n_1=2$. 
    $\frac{1}{\lambda_{He}} = R(2)^2 \left(\frac{1}{2^2} - \frac{1}{4^2}ight) = 4R\left(\frac{1}{4} - \frac{1}{16}ight) = 4R\left(\frac{3}{16}ight) = \frac{3R}{4}$.

3.  **Ratio:** $\frac{\lambda_{He}}{\lambda_H} = \frac{5R/36}{3R/4} = \frac{5}{36} 	imes \frac{4}{3} = \frac{5}{27}$.
    $\lambda_{He} = 6561 	imes \frac{5}{27} = 243 	imes 5 = 1215 \mathring{A}$.

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