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(A) The wavelength $\lambda$ for the hydrogen spectrum is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2

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$3648$

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(A) The wavelength $\lambda$ for the hydrogen spectrum is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Lyman series,$n_1 = 1$. For the shortest wavelength,$n_2 = \infty$. Thus,$\frac{1}{\lambda_L} = R(1 - 0) = R$,so $\lambda_L = \frac{1}{R} = 912 \ \mathring{A}$.For the Balmer series,$n_1 = 2$. For the shortest wavelength,$n_2 = \infty$. Thus,$\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - 0 ight) = \frac{R}{4}$.This gives $\lambda_B = \frac{4}{R} = 4 	imes \lambda_L$.Substituting the value: $\lambda_B = 4 	imes 912 \ \mathring{A} = 3648 \ \mathring{A}$.

The shortest wavelength in the Lyman series of the hydrogen spectrum is $912 \ \mathring{A}$,corresponding to a photon energy of $13.6 \ eV$. The shortest wavelength in the Balmer series is about..... $\mathring{A}$.

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