The wavelength of the first line of the Balme | vedclass

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Question

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

Vedclass · Accepted Answer

$\lambda / 9$

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(B) The Rydberg formula for the wavelength of a spectral line is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Balmer series,the first line corresponds to the transition from $n_2 = 3$ to $n_1 = 2$.Since the transition is the same for both atoms,the term $\left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$ is constant.Therefore,$\frac{1}{\lambda} \propto Z^2$,which implies $\lambda \propto \frac{1}{Z^2}$.For hydrogen $(Z_H = 1)$,$\lambda_H = \lambda$.For doubly ionized lithium $(Z_{Li} = 3)$,$\lambda_{Li} = \lambda_H 	imes \left( \frac{Z_H}{Z_{Li}} ight)^2$.$\lambda_{Li} = \lambda 	imes \left( \frac{1}{3} ight)^2 = \frac{\lambda}{9}$.

The wavelength of the first line of the Balmer series of a hydrogen atom is $\lambda \ \mathring{A}$. The wavelength of the same line for a doubly ionized lithium atom $(Z = 3)$ is:

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