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If electron falls from $n=3$ to $n=2$, then emitted energy is
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Verified Answer
The correct answer is:
$1.9 \mathrm{eV}$
$$
\Delta E=E_3-E_2=13.6\left[\frac{1}{(2)^2}-\frac{1}{(3)^2}\right]=1.9 \mathrm{eV}
$$
\Delta E=E_3-E_2=13.6\left[\frac{1}{(2)^2}-\frac{1}{(3)^2}\right]=1.9 \mathrm{eV}
$$
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