What is the wavelength of the least energetic | vedclass

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(B) The Lyman series corresponds to transitions from higher energy levels to the ground state $(n_f = 1)$.The energy of the emitted photon is given by

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$122$

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(B) The Lyman series corresponds to transitions from higher energy levels to the ground state $(n_f = 1)$.The energy of the emitted photon is given by $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight) 	ext{ eV}$.The least energetic photon corresponds to the smallest energy difference,which occurs for the transition from the first excited state $(n_i = 2)$ to the ground state $(n_f = 1)$.Using the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4}$.Thus,$\lambda = \frac{4}{3R}$.Given $R \approx 1.097 	imes 10^7 	ext{ m}^{-1}$,we have $\lambda = \frac{4}{3 	imes 1.097 	imes 10^7} \approx 1.216 	imes 10^{-7} 	ext{ m} = 121.6 	ext{ nm}$.Rounding to the nearest integer,we get $122 	ext{ nm}$.

What is the wavelength of the least energetic photon emitted in the Lyman series of the hydrogen atom spectrum? (in $nm$)

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