The ratio of energies of photons produced due | vedclass

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Question

(B) The energy of a photon emitted during a transition from energy level $n_2$ to $n_1$ is given by $\Delta E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}

Vedclass · Accepted Answer

$5:4$

Vedclass · Answer

(B) The energy of a photon emitted during a transition from energy level $n_2$ to $n_1$ is given by $\Delta E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) 	ext{ eV}$.For transition $(i)$ from $n_2 = 3$ to $n_1 = 2$:$\Delta E_1 = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = 13.6 \left( \frac{1}{4} - \frac{1}{9} ight) = 13.6 \left( \frac{5}{36} ight) 	ext{ eV}$.For transition (ii) from $n_2 = \infty$ to $n_1 = 3$:$\Delta E_2 = 13.6 \left( \frac{1}{3^2} - \frac{1}{\infty^2} ight) = 13.6 \left( \frac{1}{9} - 0 ight) = 13.6 \left( \frac{1}{9} ight) 	ext{ eV}$.The ratio is $\frac{\Delta E_1}{\Delta E_2} = \frac{13.6 	imes (5/36)}{13.6 	imes (1/9)} = \frac{5}{36} 	imes 9 = \frac{5}{4}$.

The ratio of energies of photons produced due to the transition of an electron in a hydrogen atom from its $(i)$ $3^{\text{rd}}$ to $2^{\text{nd}}$ energy level and (ii) highest energy level to $3^{\text{rd}}$ level is:

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