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In Franck-Hertz experiment, the first dip in the current-voltage graph for hydrogen is observed at 10.2 V. The wavelength of light emitted by hydrogen atom when excited to the first excitation level is _________ $\mathrm{nm}$. (Given $\mathrm{hc}=1245 \mathrm{eV} \mathrm{nm}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ ).
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The correct answer is:
122
$\begin{aligned} & 10.2 \mathrm{eV}=\frac{\mathrm{hc}}{\lambda} \\ & \lambda=\frac{1245 \mathrm{eV}-\mathrm{nm}}{10.2 \mathrm{eV}}=122.06 \mathrm{~nm}\end{aligned}$
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