Atoms & Nuclei

JEE Main level, numerical answer, medium

104 of 239

Question 104 of 239

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 _________ nm\mathrm{nm}. (Given hc=1245eVnm,e=1.6×10−19C\mathrm{hc}=1245 \mathrm{eV} \mathrm{nm}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C} ).

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Answer: 122

10.2eV=hcλλ=1245eV−nm10.2eV=122.06 nm\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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