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I am studying quantum physics and I have a question: what is the physical explanation for electrons having less energy than photons with the same wavelength?

Energy of a photon : $E = h c/\lambda$.
Energy of an electron: $E = h^2/(2m\lambda^2)$

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up vote 10 down vote accepted

For the photon we have $$E_\gamma = \frac{hc}{\lambda}$$ and for the electron $$E_e = \frac{h^2}{2m\lambda^2} =\frac{hc}{\lambda} \frac{h}{2mc\lambda} = E_\gamma \frac{h}{2mc\lambda}. $$ You can check that the proportionality factor is dimensionless. So what you are asking is why this quantity is less than unity. But recall that $$\frac{h}{\lambda} =p$$ where $p$ is the momentum. What we are looking at is really (one half) the ratio $$\frac{pc}{mc^2} = \frac{mvc}{mc^2}$$ where I assumed that $v \ll c$, that is, we have a non-relativistic electron. Then we get the result you stated in your question. On the other, hand if we don't make this approximation we have the ratio $$\frac{pc}{mc^2} =\frac{mv\gamma c}{mc^2} = \frac{v\gamma}{c}$$ which is unbounded when $v \to c$.

You could also argue from Einstein's $$E^2 = m^2 + p^2$$ (in units where $c = 1$). For $m = 0$ we have of course $E = p$. If you make a Taylor expansion of $E$ for $m\neq 0$, $$E = m + \frac{p^2}{2m} + \ldots$$ you see that the kinetic energy, compared to the energy of a massless particle has a factor $p/m$ (as we found above). The non-relativistic regime is precisely when this quantity is small, and if it is not, we have to include terms proportional to $p^4/m^3$ and higher, and again that the energy can be larger for a massive particle than for a massless particle with the same momentum. So the answer to your question really is: because you are considering non-relativistic particles.

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The energy of the particle is proportional to the oscillation frequency of its wavefunction, $E=h\nu$. A photon always moves at the speed $c$, so its wavelength is related to the frequency in the usual way for a traveling wave, $\lambda = c/\nu = hc/E$.

A massive particle moves more slowly than the photon, so its wavelength is shorter for the same amount of energy. Naively, we might guess that a particle moving at speed $v$ would have $\lambda = hv/E$ as its wavelength. This is not correct because it fails to account for relativity, but it may give you an idea of why the wavelength is shorter for a particle with mass.

To get the correct relationship, we need to consider the relativistic energy of the particle. According to special relativity, the energy is actually $E = \sqrt{p^2c^2 + m^2c^4}$. For a particle at rest, this is the famous $E=mc^2$. The kinetic energy is the difference between the total energy and the energy at rest (mass energy).

For a photon, all of the energy is kinetic because it has no mass. For a non-relativistic electron, with momentum $p \ll mc$, we can use a Taylor expansion to get an approximate expression for the kinetic energy.

\begin{align} KE &= \sqrt{m^2c^4 + p^2c^2}-mc^2\\ &\approx mc^2\left(1+{1 \over 2}{p^2c^2 \over m^2c^4 }\right)-mc^2\\ &={p^2 \over 2m} \end{align}

The DeBroglie wavelength is related to momentum by $\lambda =h/p$, and pluggin it in we obtain the formulas you asked about.

\begin{align} E_{\mathrm{photon}} &= {hc \over \lambda}\\ E_\mathrm{electron} &= {h^2 \over 2m\lambda^2} \end{align}

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A slightly different viewpoint is the following. First, note that the formulae given by the OP do not in themselves exclude the possibility that the electron has the same or greater energy than a photon of the same wavelength. Indeed, for small enough $\lambda$ the non-relativistic formula $E = h^2/(2m \lambda^2)$ predicts that the electron has greater energy than the photon. The "critical" wavelength $\lambda_c$ where the crossover happens is found by equating the two expressions, giving $$ \frac{hc}{\lambda_c} = \frac{h^2}{2m\lambda_c^2} \quad \Longrightarrow \quad \lambda_c \approx \frac{h}{mc},$$ ignoring the factor of 2. But this quantity has another meaning: it is the Compton wavelength. This is the wavelength at which the electron kinetic energy is approximately equal to its rest mass: $$ \frac{h^2}{2m\lambda_c^2} \approx m c^2. $$ Above this energy, the spontaneous production of electron-positron pairs starts to become important. Therefore, the non-relativistic concept of a "single electron" loses its meaning once the energy becomes comparable to a photon of the same wavelength.

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