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(A) The transition from $n_{i}=4$ to $n_{f}=2$ belongs to the Balmer series.The Rydberg formula for the wavenumber $(\bar{
u})$ is given by:$\bar{
u

Vedclass · Accepted Answer

$486$

Vedclass · Answer

(A) The transition from $n_{i}=4$ to $n_{f}=2$ belongs to the Balmer series.The Rydberg formula for the wavenumber $(\bar{
u})$ is given by:$\bar{
u} = \frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$Where $R_H = 1.097 	imes 10^7 \ m^{-1}$,$n_f = 2$,and $n_i = 4$.$\frac{1}{\lambda} = 1.097 	imes 10^7 \left( \frac{1}{2^2} - \frac{1}{4^2} ight)$$\frac{1}{\lambda} = 1.097 	imes 10^7 \left( \frac{1}{4} - \frac{1}{16} ight)$$\frac{1}{\lambda} = 1.097 	imes 10^7 \left( \frac{4-1}{16} ight) = 1.097 	imes 10^7 	imes \frac{3}{16}$$\frac{1}{\lambda} = 2.056875 	imes 10^6 \ m^{-1}$$\lambda = \frac{1}{2.056875 	imes 10^6} \approx 4.863 	imes 10^{-7} \ m$$\lambda = 486.3 	imes 10^{-9} \ m = 486 \ nm$.

What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with $n=4$ to an energy level with $n=2$ (in $nm$)?

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