The wavelength of the first line of Balmer se | vedclass

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Question

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

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$\frac{\lambda}{9}$

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(B) The Rydberg formula for the wavelength of a spectral line is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the first line of the Balmer series,$n_1 = 2$ and $n_2 = 3$.For hydrogen atom $(H)$,$Z = 1$. Thus,$\frac{1}{\lambda} = R (1)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right)$.For doubly ionized lithium $(Li^{2+})$,$Z = 3$. Let the wavelength be $\lambda'$.Then,$\frac{1}{\lambda'} = R (3)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 9 R \left( \frac{5}{36} \right)$.Comparing the two equations,$\frac{1}{\lambda'} = 9 \left( \frac{1}{\lambda} \right)$,which implies $\lambda' = \frac{\lambda}{9}$.

The wavelength of the first line of Balmer series of hydrogen atom is $\lambda$. What will be the wavelength of the same line in doubly ionized lithium?

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