Are protons bigger than electrons? In every text/ physics book that I've read, Protons are mentioned as particles that are bigger, way bigger 2000 times to be precise, than electrons...I believed that until a few minutes ago when I googled "what is the radius of an electron" and then I read somewhere that it was 2.5 times bigger than the radius of a PROTON...the radius of an ELECTRON is bigger than that of a PROTON. Which goes completely against every physics text book that I've read...Any help towards explaining why protons are 2000 times bigger while its radius is 2.5 times smaller than that of an electron will be valued...thanx in advance
 A: Quantum mechanical particles have well-defined masses, but they do not have well-defined sizes (radius, volume, etc) in the classical sense.  There are multiple ways you could assign a length scale to a particle, but if you think of them as little balls with a well-defined size and shape, then you're making a mistake.
de Broglie Wavelength: Particles which pass through small openings exhibit wavelike behavior, with a characteristic wavelength given by
$$\lambda_{dB} = \frac{h}{mv}$$
where $h$ is Planck's constant, $m$ is the particle's mass, and $v$ is the particle's velocity.  This sets the length scale at which quantum effects like diffraction and interference become important.  It also turns out that if the average spacing between particles in an ideal gas is on the order of $\lambda_{dB}$ or smaller, classical statistical mechanics breaks down (e.g. the entropy diverges to $-\infty$).
Compton Wavelength: One way to measure the position of a particle is to shine a laser on the region where you think the particle will be.  If a photon scatters off of the particle, you can detect the photon and trace it's trajectory back to determine where the particle was.  The resolution of a measurement like this is limited to the wavelength of the photon used, so smaller wavelength photons yield more precise measurements.
However, at a certain point the photon's energy would be equal to the mass energy of the particle.  The wavelength of such a photon is given by
$$\lambda_c = \frac{hc}{mc^2} = \frac{h}{mc}$$
Beyond this scale, position measurement stops being more precise because the photon-particle collisions start to produce particle-antiparticle pairs.
"Classical" Radius: If you want to compress a total amount of electric charge $q$ into a sphere of radius $r$, it takes energy roughly equal to $U = \frac{q^2}{4\pi\epsilon_0 r}$ (this is off by a factor of 3/5, but nevermind - we're just looking at orders of magnitude).  If we set that equal to the rest energy $mc^2$ of a (charged) particle, we find
$$r_0 = \frac{q^2}{4\pi\epsilon_0 mc^2}$$
This is sometimes called the classical radius of a particle with charge $q$ and mass $m$.  It turns out that this is of the same order of magnitude as the Thompson scattering cross section, and so this length scale is relevant when considering the scattering of low-energy electromagnetic waves off of particles.
Charge Radius: If you model a particle as a spherical "cloud" of electric charge, then you can perform very high precision scattering experiments (among other things) to determine what effective size this charge cloud has.  The result is called the charge radius of the particle, and is a very relevant length scale to consider if you are thinking about the fine details of how the particle interacts electromagnetically.  Fundamentally, the charge radius arises in composite particles because their charged constituents occupy a non-zero region of space.  The charge radius of the proton is due to the quarks of which it is comprised, and has been measured to be approximately $0.8$ femtometers; on the other hand, the electron is not known to be a composite particle, so its charge radius would be zero (which is consistent with measurements).
Excitation Energy: Yet another length scale is given by the wavelength of the photon whose energy is sufficient to excite the internal constituents of the particle into a higher energy state (e.g. of vibration or rotation).  The electron is (as far as we know) elementary, meaning that it doesn't have any constituents to excite; as a result, the electron size is zero by this measure as well.  On the other hand, the proton can be excited into a Delta baryon by a photon with energy $E\approx 300$ MeV, corresponding to a size
$$\lambda = \frac{hc}{E} \approx 4\text{ femtometers}$$

In the first three examples, note that the mass of the particle appears in the denominator; this implies that, all other things being equal, more massive particles will correspond to smaller length scales (at least by these measures).  The mass of a proton is unambiguously larger than that of an electron by a factor of approximately 1,836.  As a result, the de Broglie wavelength, Compton wavelength, and classical radius of the proton are smaller than those of the electron by the same factor.  This raises the question of where the meager 2.5x claim came from.
A quick google search shows that this claim appears on the site AlternativePhysics.org.  The point being made is that the classical electron radius mentioned above is 2.5 times the "measured" proton radius - by which they mean the measured proton charge radius.  This is true, but not particularly meaningful - being quantum mechanical objects, neither the electron nor the proton have a radius in the sense that a classical marble does.  Comparing two particles by using two completely different measures of size is comparing apples to oranges.

As a final note, I would caution you from taking any of the claims you find on AlternativePhysics.org too seriously.  To borrow a saying from the medical community, there's a name for the subset of "alternative physics" which actually makes sense.  It's called physics.
A: Reading the good last answer by Roger, it is also important to also note that an atom does not have a well-defined volume.
Treating the electron and proton as perfect spheres with even mass-density is not correct. Having said that, note
that while classical measurements may put the electron at about 2.5 times the diameter of a proton as you have stated (and a citation to that would be nice too - are you referring to the classical electron radius?),
the mass of a proton is 2000 times that of an electron.
The mass of an
electron is $\approx 9.1 \times 10^{-31} kg$ while that of the proton is $\approx 1.67 \times 10^{-27} kg$. "Size" and mass are not the same.
A: A proton is a composite particle with a radius of about 0.8-0.9 femtometers. This value is obtained from scattering and spectroscopic data that are sensitive to the details of the coulomb potential at very small scale.
For all we know an electron is a point particle. No internal degrees of freedom besides spin were found and scattering data are consistent with an upper limit for the radius of $10^{-18}$ m (from wikipedia but with a broken link as reference). The unsolved issue is that the EM self energy diverges for a point particle. For a radius of 2.8 femtometers this self energy is already equal to the electron mass, which is why this value is known as the (Thomson) radius of the electron. It is this number that caused your confusion.
A: The fact behind this claim is that the masses of protons and neutrons are about 2000 times greater than those of electrons. The mass is more objective and permanent characteristic of a particle than its size (which is often defined as the extent of its wave function and can vary significantly in various circumstances).
A: Let me give you the crazy idea that the radius of an electron and a proton is fixed but complex, where the real part is the mean and the imaginary part is the standard deviation. Then the classical radius of an electron and a proton determines the mean value, and the root-mean-square value is variable in its meaning. The electron radius is pointwise at high energies, when relativistic corrections are applied, and the scattering cross section is proportional to the square off classical electron radius.
The formula for the scattering cross section of a photon by an electron does not need to be regularized and determines the scattering cross section
$$Re\sigma=\sigma(0)-\sigma(\infty)=\frac{8}{3}\pi r_e^2;\sigma(x)=\sigma(\frac{\hbar \omega}{mc^2})$$
In this case, the radius in complex form is
$$R_e=r_e(1\pm\sqrt{(Re\sigma-\pi r_e^2)/\pi}i)=r_e(1\pm 1.29i)$$
its modulus determines the scattering cross section
$$|R_e|=r_e|1\pm1.29i|=1.63r_e=\sqrt{\frac{8}{3}}r_e$$
The formulas for the cross section of the scattering of an electron by an electron and the annihilation of an electron and a positron with the formation of two photons require regularization. The regularization parameter must be chosen so that the size of the electron coincides with the size of the electron when a photon is scattered by an electron. It turns out that the three formulas equally determine the size of the electron.
There is no unambiguous value for the size of elementary particles. Elementary particles do not have a finite size and it is impossible to determine an unambiguous final size by their charge. For an electron, there are scattering cross-sections of various reactions, and with their help I was able to determine the complex size of an electron. The complex size of an electron is determined up to the imaginary part. For a proton, this cannot be done, since there are no formulas describing the cross-sectional area of ​​reactions. Nuclear forces are not described by the perturbation theory, therefore only measurements are made and there are no theoretical formulas. The classical radius of the electron is greater than the classical radius of the proton. But this does not mean anything, the size of the proton is unknown.
