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(B) The Rydberg formula for the hydrogen atom $(Z=1)$ is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.The ultravi

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$823$

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(B) The Rydberg formula for the hydrogen atom $(Z=1)$ is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$.The ultraviolet region corresponds to the Lyman series $(n_f = 1)$. The largest wavelength corresponds to the minimum energy transition,which is from $n_i = 2$ to $n_f = 1$.Given $\lambda_{max, UV} = 122 \ nm$,we have $\frac{1}{122} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4}$.Thus,$\frac{1}{R} = 122 	imes \frac{3}{4} = 91.5 \ nm$.The infrared region includes the Paschen series $(n_f = 3)$,Brackett series $(n_f = 4)$,and Pfund series $(n_f = 5)$. The smallest wavelength in the entire infrared region corresponds to the maximum energy transition,which is the limit of the Paschen series ($n_f = 3$ to $n_i = \infty$).$\frac{1}{\lambda_{min, IR}} = R \left( \frac{1}{3^2} - \frac{1}{\infty^2} ight) = \frac{R}{9}$.$\lambda_{min, IR} = \frac{9}{R} = 9 	imes 91.5 = 823.5 \ nm$.Rounding to the nearest integer,we get $823 \ nm$. Therefore,option $(B)$ is correct.

The largest wavelength in the ultraviolet region of the hydrogen spectrum is $122 \ nm$. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is (in $nm$)

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