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(D) For a hydrogen atom,the Rydberg formula is given by: $\frac{1}{\lambda} = R_H 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.Here,$n_1 = 1

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$91 \ nm$

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(D) For a hydrogen atom,the Rydberg formula is given by: $\frac{1}{\lambda} = R_H 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.Here,$n_1 = 1$ (first stationary state),$n_2 = \infty$ (infinite state),$Z = 1$ (for hydrogen),and $R_H = 1.097 	imes 10^7 \ m^{-1}$.Substituting the values: $\frac{1}{\lambda} = 1.097 	imes 10^7 	imes 1^2 	imes (\frac{1}{1^2} - \frac{1}{\infty^2})$.$\frac{1}{\lambda} = 1.097 	imes 10^7 	imes (1 - 0) = 1.097 	imes 10^7 \ m^{-1}$.$\lambda = \frac{1}{1.097 	imes 10^7} \approx 9.116 	imes 10^{-8} \ m = 91.16 \ nm$.Thus,the wavelength is approximately $91 \ nm$.

When an electron in a hydrogen atom jumps from an infinite state to the first stationary state,the wavelength of the emitted radiation will be ...... (Rydberg constant = $1.097 \times 10^7 \ m^{-1}$):

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