If an electron falls from n = 3 to n = 2,then | vedclass

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Question

(C) The energy emitted when an electron transitions between energy levels in a hydrogen atom is given by the formula: $\Delta E = 13.6 \, 	ext{eV}

Vedclass · Accepted Answer

$1.9$

Vedclass · Answer

(C) The energy emitted when an electron transitions between energy levels in a hydrogen atom is given by the formula: $\Delta E = 13.6 \, 	ext{eV} 	imes \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$.Given $n_i = 3$ and $n_f = 2$,we have:$\Delta E = 13.6 	imes \left( \frac{1}{2^2} - \frac{1}{3^2} ight) \, 	ext{eV}$.$\Delta E = 13.6 	imes \left( \frac{1}{4} - \frac{1}{9} ight) \, 	ext{eV}$.$\Delta E = 13.6 	imes \left( \frac{9 - 4}{36} ight) \, 	ext{eV}$.$\Delta E = 13.6 	imes \frac{5}{36} \, 	ext{eV} \approx 1.89 \, 	ext{eV}$,which rounds to $1.9 \, 	ext{eV}$.

If an electron falls from $n = 3$ to $n = 2$,then the emitted energy is $..........$ $eV$.

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