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(A) For the Lyman series,the transition occurs to the ground state,so $n_1 = 1$. The electron transitions from the $4^{th}$ orbit,so $n_2 = 4$.Using t

Vedclass · Accepted Answer

$9.7 	imes 10^{-8} \ m$

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(A) For the Lyman series,the transition occurs to the ground state,so $n_1 = 1$. The electron transitions from the $4^{th}$ orbit,so $n_2 = 4$.Using the Rydberg formula: $\frac{1}{\lambda} = R_H 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2})$Given $R_H = 1.097 	imes 10^7 \ m^{-1}$ and $Z = 1$ for hydrogen:$\frac{1}{\lambda} = 1.097 	imes 10^7 	imes 1^2 	imes (\frac{1}{1^2} - \frac{1}{4^2})$$\frac{1}{\lambda} = 1.097 	imes 10^7 	imes (1 - \frac{1}{16})$$\frac{1}{\lambda} = 1.097 	imes 10^7 	imes \frac{15}{16} \ m^{-1}$$\lambda = \frac{16}{1.097 	imes 10^7 	imes 15} \ m$$\lambda \approx 9.7 	imes 10^{-8} \ m$

If an electron in a hydrogen atom undergoes a transition from the $4^{th}$ stationary orbit and emits a line in the Lyman series,calculate the wavelength of the emitted radiation.

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