Why can't the electron enter the nucleus? Actually, there exists an attractive force between an electron and the protons inside the nucleus, but the electron cannot be attracted towards the nucleus! What force balances that attractive force? Explain.
 A: The probability distribution for finding a ground-state hydrogen atom's electron
in some volume is given by $dP = |\psi|^2 d^3x$, where the wavefunction $\psi$
is given by
$$
\psi_{n\ell m} = \psi_{000} = \frac1{\sqrt{4\pi}}\frac2{a_0^{3/2}}e^{-r/a_0}
$$
where $a_0 \approx \frac12\times10^{-10}\rm\,m$ is the Bohr radius.  This is the first of the wavefunctions plotted by Sofia above.
The radius of the nucleus is, charitably, $r_n \approx \frac32\times10^{-15}\rm\,m$, about five orders of magnitude smaller.
We compute the probability of finding the electron in the nucleus by doing the integral over the nuclear volume,
$$
P = 4\pi\int_{r=0}^{r_n}r^2 dr |\psi|^2 
\approx \frac43\left(
\frac{r_n}{a_0}
\right)^3
$$
which gives a roughly $10\times10^{-15}$, with most of the uncertainty coming from my guess for $r_n$.  So if you had a gram of hydrogen atoms, about ten femtograms of them have the electron overlapping the nucleus at any instant.
The thing is that the electron occupies the entire volume of the atom, while the nucleus is astoundingly smaller.  If we didn't have electron capture, we wouldn't talk about electrons spending time within the nucleus at all.
A: The reason is there is no known force that is strong enough to hold it there. Electrons are quantum particles having very small mass. But we can show how order of magnitude calculations using a minimum amount of quantum mechanics (the position-momentum uncertainty principle) and mechanical energy principles lead to correct order of magnitude results for the hydrogen atom in its ground state.
Consider an electron in the coulomb field of a proton, which is assumed to be stationary at the origin of the coordinate system. If the two particles are separated by a distance $r$, the potential energy of the electron is
$V(r) = \frac{-q^2}{4\pi\epsilon_0}\frac{1}{r}$
where $q$ is its electric charge, exactly opposite the charge of the proton. Set 
$\frac{q^2}{4\pi\epsilon_0} = e^2$
Assume that the state of the electron is described by a spherically symmetric wave function whose spatial extent is characterized by $r_0$. The potential energy corresponding to this state is then on the order of
$\bar V \simeq \frac{-e^2}{r_0}$
For it to be as low as possible, it is necessary to have $r_0$ as small as possible. That is, the wave function must be as concentrated as possible about the proton. But we must also take the kinetic energy into account. This is where the uncertainty principle comes in: if the electron is confined within a volume of linear dimension $r_0$, the uncertainty $\Delta p$ in its momentumis at least of the order of $\frac{\hbar}{r_0}$. In other words, even if the average momentum is zero, the kinetic energy $T$ associated with the state under consideration is not zero:
$\bar T \gtrsim \bar T_{min} = \frac{1}{2m}(\Delta p)^2 \simeq \frac {\hbar^2}{2m(r_0)^2}$
If we take  $r_0$ smaller in order to decrease the potential energy, the minimum kinetic energy increases.
The Lowest total energy compatible with the uncertainty relation is thus the minimum of the function:
$E_{min} = \bar T_{min} + \bar V = \frac {\hbar^2}{2m{r_0}^2} - \frac{e^2}{r_0}$
This minimum is obtained for:
$r_0 = a_0 = \frac{\hbar^2}{me^2}$
and is equal to:
$E_0 = -\frac{me^4}{2\hbar^2}$
Because of the uncertainty relation, the smaller the extension of the wave function the greater the kinetic energy of the electron. You could say the uncertainty relation provides a buoyancy that lifts the electron off of the proton in the nucleus.
Source: Quantum Mechanics Volume 1, Cohen-Tannoudji, Diu and Laloe, John Wiley and sons, 1977
A: The electron is attracted by the nucleus. In the Bohr model it was said that the electron rotates around the nucleus and the centrifugal force compensates the nuclear attraction. But the Bohr model is outdated. The electron is a quantum particle in a electrostatic field, and its behavior is dictated by the Schrodinger equation and ultimately by the wave-function, see the picture. 
As you can see from the left side column in the picture, in the state $n=1$, i.e. on the inner electron shell, the electron has a non-zero and non-negligible probability to be found at the nucleus position - see the innermost bright region. Moreover, the phenomenon of electron-capture by a nucleus is known, see here.
"Electron capture is a process in which a proton-rich nuclide absorbs an inner atomic electron, thereby changing a nuclear proton to a neutron and simultaneously causing the emission of an electron neutrino. Various photon emissions follow, as the energy of the atom falls to the ground state of the new nuclide."
But the biggest probability is that the electron be outside the nucleus. The radius of the electron cloud is $10^{-8}$cm, and the radius of the nucleus is cca. $10^{-12}$cm.

A: Electrons are like planets revolving around a nucleus.
As centrifugal force and cetripetal force had same magnitudes.
When the netforce is 0 from all directions the particle will start spinning.The same happens for electron and start revolving around the nucleus.
