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

Vedclass · Accepted Answer

$1215 \mathring A$

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(B) The Rydberg formula for the wavelength of spectral lines is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) Z^2$.For the Lyman series,the electron transitions to the ground state,so $n_1 = 1$.The first spectral line of the Lyman series corresponds to the transition from the first excited state to the ground state,so $n_2 = 2$.For hydrogen,the atomic number $Z = 1$.Substituting these values into the formula:$\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) (1)^2 = R \left( 1 - \frac{1}{4} ight) = R \left( \frac{3}{4} ight)$.Therefore,$\lambda = \frac{4}{3R}$.Given that $\frac{1}{R} \approx 911.6 \mathring A$ (often approximated as $912 \mathring A$),we have:$\lambda = \frac{4}{3} 	imes 911.6 \mathring A \approx 1215.5 \mathring A$.Rounding to the nearest option,the wavelength is $1215 \mathring A$.

The wavelength of the first spectral line of the Lyman series of the hydrogen spectrum is

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