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(C) The Rydberg formula for the Balmer series is given by $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$,where $n = 3, 4, 5, \do

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$\frac{9}{\lambda_1}-\frac{9}{\lambda_2}$

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(C) The Rydberg formula for the Balmer series is given by $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$,where $n = 3, 4, 5, \dots$. For the shortest wavelength $\lambda_1$,$n = \infty$: $\frac{1}{\lambda_1} = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4} \implies R = \frac{4}{\lambda_1}$. For the longest wavelength $\lambda_2$,$n = 3$: $\frac{1}{\lambda_2} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right) \implies R = \frac{36}{5\lambda_2}$. To express $R$ in terms of $\lambda_1$ and $\lambda_2$,we use the relation: $\frac{1}{\lambda_1} = \frac{R}{4}$ and $\frac{1}{\lambda_2} = \frac{5R}{36}$. Multiplying the first by $9$: $\frac{9}{\lambda_1} = \frac{9R}{4}$. Subtracting the second: $\frac{9}{\lambda_1} - \frac{1}{\lambda_2} = \frac{9R}{4} - \frac{5R}{36} = \frac{81R - 5R}{36} = \frac{76R}{36} = \frac{19R}{9}$. Alternatively,using the derived values: From $\frac{1}{\lambda_1} = \frac{R}{4}$,$R = \frac{4}{\lambda_1}$. From $\frac{1}{\lambda_2} = \frac{5R}{36}$,$R = \frac{36}{5\lambda_2}$. Equating or manipulating the options,we find that for $R = \frac{9}{\lambda_1} - \frac{9}{\lambda_2}$,we get $R = 9(\frac{R}{4} - \frac{5R}{36}) = 9(\frac{9R-5R}{36}) = 9(\frac{4R}{36}) = R$. Thus,the correct expression is $R = \frac{9}{\lambda_1} - \frac{9}{\lambda_2}$.

In the hydrogen spectrum,the shortest and longest wavelengths of the Balmer series are $\lambda_1$ and $\lambda_2$ respectively. The Rydberg constant $R$ of hydrogen is:

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