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(B) For a hydrogen atom,$Z = 1$. The Rydberg formula is $\frac{1}{\lambda} = R_H 	imes (1)^2 	imes \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$

Vedclass · Accepted Answer

$486$

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(B) For a hydrogen atom,$Z = 1$. The Rydberg formula is $\frac{1}{\lambda} = R_H 	imes (1)^2 	imes \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.Given $n_1 = 2$ and $n_2 = 4$,and $R_H = 1.097 	imes 10^7 \ m^{-1}$.$\frac{1}{\lambda} = 1.097 	imes 10^7 	imes \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = 1.097 	imes 10^7 	imes \left( \frac{1}{4} - \frac{1}{16} ight)$.$\frac{1}{\lambda} = 1.097 	imes 10^7 	imes \left( \frac{3}{16} ight) = 2.0568 	imes 10^6 \ m^{-1}$.$\lambda = \frac{1}{2.0568 	imes 10^6} \approx 4.86 	imes 10^{-7} \ m = 486 \ nm$.

The wavelength of radiation emitted when an electron falls from the $4^{th}$ Bohr orbit to the $2^{nd}$ in an $H$ atom is .............. $nm$.

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