If the wavelength of the first line of the Pa | vedclass

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(C) The Rydberg formula for the hydrogen atom is $\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$.For the Paschen series,$n_1

Vedclass · Accepted Answer

$492$

Vedclass · Answer

(C) The Rydberg formula for the hydrogen atom is $\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}ight)$.For the Paschen series,$n_1 = 3$. The first line corresponds to $n_2 = 4$,and the second line corresponds to $n_2 = 5$.For the first line: $\frac{1}{\lambda_1} = R_H \left(\frac{1}{3^2} - \frac{1}{4^2}ight) = R_H \left(\frac{1}{9} - \frac{1}{16}ight) = R_H \left(\frac{16-9}{144}ight) = R_H \left(\frac{7}{144}ight)$.Given $\lambda_1 = 720 \ nm$,so $\frac{1}{720} = R_H \left(\frac{7}{144}ight)$.For the second line: $\frac{1}{\lambda_2} = R_H \left(\frac{1}{3^2} - \frac{1}{5^2}ight) = R_H \left(\frac{1}{9} - \frac{1}{25}ight) = R_H \left(\frac{25-9}{225}ight) = R_H \left(\frac{16}{225}ight)$.Taking the ratio: $\frac{\lambda_1}{\lambda_2} = \frac{R_H (16/225)}{R_H (7/144)} = \frac{16}{225} 	imes \frac{144}{7} = \frac{2304}{1575}$.$\lambda_2 = \lambda_1 	imes \frac{1575}{2304} = 720 	imes \frac{1575}{2304} = 492.1875 \ nm$.The nearest integer is $492 \ nm$.

If the wavelength of the first line of the Paschen series of a hydrogen atom is $720 \ nm$,then the wavelength of the second line of this series is $.......... \ nm$. (Nearest integer)

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