In an electron transition from an infinite st | vedclass

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(A) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = R \left[\frac{1}{n_1^2} - \frac{1}{n_2^2}\right]$For

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$91$

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(A) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = R \left[\frac{1}{n_1^2} - \frac{1}{n_2^2}ight]$For the transition from the infinite state $(n_2 = \infty)$ to the ground state $(n_1 = 1)$ in an $H$ atom:$R = 1.097 	imes 10^7 \ m^{-1}$Substituting the values: $\frac{1}{\lambda} = 1.097 	imes 10^7 \left[\frac{1}{1^2} - \frac{1}{\infty^2}ight] = 1.097 	imes 10^7 \ m^{-1}$Therefore,$\lambda = \frac{1}{1.097 	imes 10^7} \approx 9.116 	imes 10^{-8} \ m = 91.16 \ nm \approx 91 \ nm$.

In an electron transition from an infinite state to the ground state in an $H$ atom,find the emitted wavelength. (in $nm$)

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