The ratio of energies of photons produced due | vedclass

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(D) The energy of a photon emitted during an electronic transition is given by $E = Rhc \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$.Here

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$3$

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(D) The energy of a photon emitted during an electronic transition is given by $E = Rhc \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$.Here,$R$ is the Rydberg constant,$h$ is Planck's constant,and $c$ is the speed of light.$(i)$ For the transition from the second $(n_2 = 2)$ to the first $(n_1 = 1)$ energy level:$E_{1} = Rhc \left( \frac{1}{1^{2}} - \frac{1}{2^{2}} \right) = Rhc \left( 1 - \frac{1}{4} \right) = \frac{3}{4} Rhc$.$(ii)$ For the transition from the highest energy level $(n_2 = \infty)$ to the second $(n_1 = 2)$ energy level:$E_{2} = Rhc \left( \frac{1}{2^{2}} - \frac{1}{\infty^{2}} \right) = Rhc \left( \frac{1}{4} - 0 \right) = \frac{1}{4} Rhc$.The ratio of the energies is $\frac{E_{1}}{E_{2}} = \frac{\frac{3}{4} Rhc}{\frac{1}{4} Rhc} = \frac{3}{1}$.Therefore,the ratio is $3: 1$.

The ratio of energies of photons produced due to the transition of an electron of a hydrogen atom from its $(i)$ second to first energy level and $(ii)$ highest energy level to $2^{nd}$ level is respectively: (in $: 1$)

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