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A photon of wavelength 300 nm interacts with a stationary hydrogen atom in ground state. During the interaction, whole energy the photon is transferred to the electron of the atom. State which possibility is correst. (Consider, Plank constant $=4 \times 10^{-15} eV_{s},$ velocity of light $=3 \times 10^{8} \mathrm{m} / \mathrm{s},$ ionisation energy of hydrogen $=13.6 \mathrm{eV}$ )
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The correct answer is:
Electron will keep orbiting in the ground state of the atom
The energy of the photon
$$
\begin{aligned}
E &=\frac{h C}{\lambda} \\
&=\frac{4 \times 10^{-15} \mathrm{eVs} \times 3 \times 10^{8} \mathrm{m} / \mathrm{s}}{300 \times 10^{-9} \mathrm{m}} \\
&=\frac{4 \times 10^{2}}{300} \mathrm{eV}=\frac{4}{3} \mathrm{eV}=1.33 \mathrm{eV}
\end{aligned}
$$
The ionisation energy is $13.6 \mathrm{eV}$ which is greater than energy of photon, so atom can not come into excited state and will remain in ground state
$$
\begin{aligned}
E &=\frac{h C}{\lambda} \\
&=\frac{4 \times 10^{-15} \mathrm{eVs} \times 3 \times 10^{8} \mathrm{m} / \mathrm{s}}{300 \times 10^{-9} \mathrm{m}} \\
&=\frac{4 \times 10^{2}}{300} \mathrm{eV}=\frac{4}{3} \mathrm{eV}=1.33 \mathrm{eV}
\end{aligned}
$$
The ionisation energy is $13.6 \mathrm{eV}$ which is greater than energy of photon, so atom can not come into excited state and will remain in ground state
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