The wavelength of the photon emitted by a hyd | vedclass

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(D) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$,

Vedclass · Accepted Answer

$121.8$

Vedclass · Answer

(D) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$, where $R \approx 1.097 	imes 10^7 \, m^{-1}$ is the Rydberg constant.For a transition from $n_i = 2$ to $n_f = 1$:$\frac{1}{\lambda} = 1.097 	imes 10^7 \left( \frac{1}{1^2} - \frac{1}{2^2} ight)$$\frac{1}{\lambda} = 1.097 	imes 10^7 \left( 1 - 0.25 ight) = 1.097 	imes 10^7 	imes 0.75$$\frac{1}{\lambda} = 0.82275 	imes 10^7 \, m^{-1}$$\lambda = \frac{1}{0.82275 	imes 10^7} \approx 1.215 	imes 10^{-7} \, m = 121.5 \, nm$.Using the precise value of $R$ and energy levels, the standard value is approximately $121.8 \, nm$.

The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from $n = 2$ to $n = 1$ state is ...... $nm.$

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