For a hydrogen (H) atom,what will be the wave | vedclass

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

Vedclass · Accepted Answer

$91$

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(C) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.Given: $n_1 = 1$ (first energy level),$n_2 = \infty$ (infinite energy level),and $R_H = 1.097 	imes 10^{7} \ m^{-1}$.Substituting the values: $\frac{1}{\lambda} = 1.097 	imes 10^{7} \left( \frac{1}{1^2} - \frac{1}{\infty^2} ight)$.Since $\frac{1}{\infty} = 0$,we have $\frac{1}{\lambda} = 1.097 	imes 10^{7} \ m^{-1}$.Therefore,$\lambda = \frac{1}{1.097 	imes 10^{7}} \ m = 0.9115 	imes 10^{-7} \ m$.Converting to nanometers $(nm)$: $\lambda = 0.9115 	imes 10^{-7} 	imes 10^{9} \ nm = 91.15 \ nm \approx 91 \ nm$.

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

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