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Question

(A) The wavelength $\lambda$ of a photon emitted during a transition between energy levels is given by the Rydberg formula: $\frac{1}{\lambda} = R Z^2

Vedclass · Accepted Answer

$4:1$

Vedclass · Answer

(A) The wavelength $\lambda$ of a photon emitted during a transition between energy levels is given by the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.Since the transition is from the same excited state $n$ to the ground state $(n_1 = 1, n_2 = n)$ for both atoms,the term $\left( \frac{1}{1^2} - \frac{1}{n^2} \right)$ is constant.Therefore,$\frac{1}{\lambda} \propto Z^2$,which implies $\lambda \propto \frac{1}{Z^2}$.For Hydrogen $(Z_H = 1)$ and Helium $(Z_{He} = 2)$:$\frac{\lambda_H}{\lambda_{He}} = \frac{Z_{He}^2}{Z_H^2} = \frac{2^2}{1^2} = \frac{4}{1}$.Thus,the ratio of the wavelengths is $4:1$.

In $(i)$ a hydrogen atom and $(ii)$ a singly ionized helium atom,an electron makes a transition from the same excited state $n$ to the ground state. The ratio of the wavelengths of the emitted photons in the two cases will be:

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