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Question

(D) The wavelength of the emitted photon for a transition from the $n^{th}$ state to the ground state is given by the Rydberg formula: $\frac{1}{\Lamb

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$\Lambda_n = A + \frac{B}{\lambda_n^2}$

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(D) The wavelength of the emitted photon for a transition from the $n^{th}$ state to the ground state is given by the Rydberg formula: $\frac{1}{\Lambda_n} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right)$.For large $n$,we can approximate this as $\Lambda_n = \frac{1}{R} (1 - \frac{1}{n^2})^{-1} \approx \frac{1}{R} (1 + \frac{1}{n^2}) = \frac{1}{R} + \frac{1}{R n^2}$.The de Broglie wavelength of an electron in the $n^{th}$ orbit is $\lambda_n = \frac{h}{p} = \frac{h}{m v_n}$. Since $v_n \propto \frac{1}{n}$,we have $\lambda_n \propto n$,which implies $n^2 \propto \lambda_n^2$.Substituting this into the expression for $\Lambda_n$,we get $\Lambda_n = A + \frac{B}{\lambda_n^2}$,where $A = \frac{1}{R}$ and $B$ is a constant related to the physical parameters of the atom.

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