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(D) The energy of a photon emitted during a transition is given by $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,a

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$\frac{1}{\lambda_1} = \frac{1}{\lambda_2} + \frac{1}{\lambda_3}$

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(D) The energy of a photon emitted during a transition is given by $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength.For the first three energy levels $(n=1, 2, 3)$,the possible transitions are:$1$. $n=3 \rightarrow n=1$ (Energy $E_1$,wavelength $\lambda_1$)$2$. $n=2 \rightarrow n=1$ (Energy $E_2$,wavelength $\lambda_2$)$3$. $n=3 \rightarrow n=2$ (Energy $E_3$,wavelength $\lambda_3$)Since energy is inversely proportional to wavelength $(E \propto \frac{1}{\lambda})$,the transition with the largest energy change corresponds to the shortest wavelength. Thus,$\lambda_1$ is the shortest wavelength $(n=3 \rightarrow n=1)$,and $\lambda_3$ is the longest wavelength $(n=3 \rightarrow n=2)$.From the conservation of energy,the energy of the transition from $n=3$ to $n=1$ is equal to the sum of the energies of the transitions from $n=3$ to $n=2$ and $n=2$ to $n=1$:$E_{3 \rightarrow 1} = E_{3 \rightarrow 2} + E_{2 \rightarrow 1}$Substituting $E = \frac{hc}{\lambda}$:$\frac{hc}{\lambda_1} = \frac{hc}{\lambda_3} + \frac{hc}{\lambda_2}$Dividing by $hc$ on both sides,we get:$\frac{1}{\lambda_1} = \frac{1}{\lambda_2} + \frac{1}{\lambda_3}$

Due to transitions among its first three energy levels,a hydrogenic atom emits radiation at three discrete wavelengths $\lambda_1, \lambda_2$ and $\lambda_3$ (where $\lambda_1 < \lambda_2 < \lambda_3$). Then,

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