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Minimum excitation potential of Bohr's first orbit in hydrogen atom is
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Verified Answer
The correct answer is:
10.2 V
From the relation
$\begin{aligned} \Delta E & =13.6 \mathrm{eV}-\frac{13.6 \mathrm{eV}}{n^2} \\ & =13.6 \mathrm{eV}-\frac{13.6 \mathrm{eV}}{(2)^2}=10.2 \mathrm{eV}\end{aligned}$
Therefore, excitation potential
$=\frac{10.2}{e} \mathrm{eV}=10.2 \mathrm{~V}$
$\begin{aligned} \Delta E & =13.6 \mathrm{eV}-\frac{13.6 \mathrm{eV}}{n^2} \\ & =13.6 \mathrm{eV}-\frac{13.6 \mathrm{eV}}{(2)^2}=10.2 \mathrm{eV}\end{aligned}$
Therefore, excitation potential
$=\frac{10.2}{e} \mathrm{eV}=10.2 \mathrm{~V}$
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