In terms of Rydberg constant R, the shortest | vedclass

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

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$\frac{4}{R}$

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(B) The Rydberg formula for the wavelength $\lambda$ 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 Balmer series,the transition ends at $n_1 = 2$.For the shortest wavelength,the transition must occur from the highest possible energy level,i.e.,$n_2 = \infty$.Substituting these values for a Hydrogen atom $(Z = 1)$: $\frac{1}{\lambda} = R(1)^2 \left[ \frac{1}{2^2} - \frac{1}{\infty^2} \right]$.Since $\frac{1}{\infty} = 0$,we get: $\frac{1}{\lambda} = R \left[ \frac{1}{4} - 0 \right] = \frac{R}{4}$.Therefore,the shortest wavelength is $\lambda = \frac{4}{R}$.

In terms of Rydberg constant $R,$ the shortest wavelength in the Balmer series of the Hydrogen atom spectrum will be:

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