For a H-atom,what is the wavelength in nm ass | vedclass

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(C) The Rydberg formula is given by: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$Here,$n_1 = 1$ (first energy level) and

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$91$

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(C) The Rydberg formula is given by: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$Here,$n_1 = 1$ (first energy level) and $n_2 = \infty$ (infinite energy level).Substituting the values: $\frac{1}{\lambda} = 1.097 	imes 10^7 \ m^{-1} \left( \frac{1}{1^2} - \frac{1}{\infty^2} ight)$Since $\frac{1}{\infty} = 0$,we have: $\frac{1}{\lambda} = 1.097 	imes 10^7 \ m^{-1}$$\lambda = \frac{1}{1.097 	imes 10^7} \ m \approx 0.9116 	imes 10^{-7} \ m = 91.16 	imes 10^{-9} \ m = 91.16 \ nm$.Rounding to the nearest integer,the wavelength is $91 \ nm$.

For a $H$-atom,what is the wavelength in $nm$ associated with the transition of an electron from an infinite energy level to the first energy level? (Rydberg constant = $1.097 \times 10^{7} \ m^{-1}$)

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