The minimum energy required to excite a hydrogen electron is 10.2 eV. When photons of energy spread over a continuous range of wavelengths fall on a sample of hydrogen, why are only those photons absorbed which have energies corresponding to exact energy differences between the allowed states? For a transition requiring energy $E=h\nu_0$, a photon with frequency $\nu>\nu_0$ can be absorbed, with the remaining energy $\Delta E=h\nu-h\nu_0$ appearing as kinetic energy of the atom. This would result in a continuous spectrum, however – with no missing frequencies.
[Physics] Why is absorption spectrum a line spectrum
atomic-physicsspectroscopy
Best Answer
What you're describing does happen to some extent. It's called "Doppler broadening": absorption and emission lines in hot materials are wider (in wavelength) than absorption and emission lines in cool materials, because atoms in the hot material occupy a broader range of velocities.
An atom at rest can't absorb any old photon and convert the extra energy into energy of motion, because such collisions must conserve both energy and momentum. Let's start with the decay of an excited atom by photon emission. An $n=1$ excited hydrogen atom at rest has energy $$E_\text{excited} = m_\text{ground}c^2 + \Delta E \approx \rm 1\,GeV + 10\,eV$$ and decays to a ground-state atom nearly at rest and a photon, with energies $$ E_\text{ground} = mc^2 + \frac{p^2}{2m} \approx \rm 1\,GeV \qquad \it E_\gamma = h\nu \approx \rm 10\,eV $$ Conservation of energy tells us that $$ \Delta E = \frac{p^2}{2m} + h\nu $$
By conservation of momentum we must have the same magnitude of momentum $p$ for the atom and for the photon. The photon's momentum is $p_\gamma = h\nu/c$, so we get $$ \Delta E = \frac{1}{2m} \left(\frac{h\nu}{c}\right)^2 + h\nu = h\nu \left( \frac{h\nu}{2mc^2} + 1 \right) \approx h\nu $$ So you can see that the atom does get a kick from emitting the photon, but it's at the parts-per-billion level of the photon's energy.
You're asking about absorption. In the center-of-mass reference frame, every absorption must look exactly like the process we've just gone through, only with the initial and final states switched. If you want your ground-state hydrogen atom to absorb a 12 eV photon, it must be "running away" from the photon so the the Doppler-shifted energy in the center-of-mass frame corresponds to 10 eV.
This is different from elastic collisions between billiard balls, because there is no photon in the final state, and different from inelastic collisions like the "ballistic pendulum" because the atom can only absorb energy from the electromagnetic field, and thus momentum from the electromagnetic field, in specially-sized lumps.
We can be even more specific. Let's start with an incident photon of unspecified energy and a ground-state atom at rest, $$ E_\text{ground} = mc^2 \qquad E_\gamma = h\nu $$ and try to end up with a "kicked" excited atom, $$ E_\text{excited} = mc^2 + \Delta E + \frac{p^2}{2m}. $$ From conservation of energy we have \begin{align} E_\text{excited} &= E_\text{ground} + E_\gamma \\ \Delta E + \frac{p^2}{2m} &= h\nu. \tag{A} \end{align} By the same logic as above we have $|p| = h\nu$ (though now the directions are parallel, rather than antiparallel, since we are not in the center-of-momentum frame), which gives \begin{align} \Delta E &= h\nu \left( 1 - \frac{h\nu}{2mc^2} \right) \\ 0 &= \Delta E - h\nu + \frac{1}{2mc^2} (h\nu)^2 \\ h\nu &= mc^2 \left( 1 \pm \sqrt{ 1 - \frac{2\Delta E}{mc^2} } \right) \end{align} This quadratic equation has exactly two solutions: $h\nu \approx 2mc^2$ (a relativistic photon interaction that we're not interested in) and the one we expected, $h\nu \approx \Delta E$. If we keep two terms in the binomial expansion for the square root we get \begin{align} \frac{h\nu}{mc^2} &\approx 1 - 1 + \frac12 \frac{2\Delta E}{mc^2} + \frac18 \left(\frac{2\Delta E}{mc^2} \right)^2 \\ h\nu & \approx \Delta E \left( 1 + \frac12 \frac{\Delta E}{mc^2} \right), \tag{B} \end{align} which is more than the "transition energy" by an extra amount $\epsilon = (\Delta E)^2/2mc^2$. That's the only extra energy that you're allowed to impart to the atom at rest, and it's required by momentum conservation and small enough to usually be negligible.