alleryAn electron in a hydrogen atom undergoes a transition from $n = 3$ to $n = 2$. The energy of the emitted photon is approximately:
A$10.2\,\text{eV}$
B$1.89\,\text{eV}$
C$0.66\,\text{eV}$
D$3.4\,\text{eV}$
Answer & Solution
Correct answer: B. $1.89\,\text{eV}$
$E_3 = -13.6/9 = -1.51\,\text{eV}$, $E_2 = -13.6/4 = -3.4\,\text{eV}$.
Energy of emitted photon $= E_3 - E_2 = -1.51 - (-3.4) = 1.89\,\text{eV}$.
This is the H$_\alpha$ line of the Balmer series, $\lambda \approx 656\,\text{nm}$ in the red.
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