The electrons cannot exist inside the nucleus because
(A) De broglie wavelength associated with electron decay is much less than the size of the nucleus
(B) De broglie wavelength associated with electron decay is much greater than the size of the nucleus
(C) De broglie wavelength associated with electron decay in equal to the size of the nucleus
(D) Negative charge cannot exist in nucleus

Hint:The de broglie wavelength by the wave particle duality is a wavelength which has been manifested in all the objects in quantum mechanics. It tends to determine the probability density of finding the object at a point given the configuration of space. The de broglie wavelength has been inversely proportional to that of momentum.

Complete step-by-step answer: We will use the Heisenberg uncertainty principle here for solving this question. The formula for this is:
$\Delta x\Delta P\underline > \dfrac{h}{{4\pi }}$
So now the approximate value for momentum will be:
$\Delta P\underline > \dfrac{h}{{4\pi \Delta x}}$
So this $\Delta x$can be equal to ${10^{ - 15}}$m as we are talking about inside the nucleus here.
So the formula for de broglie wavelength is :
$\lambda = \dfrac{h}{P}$
So we can approximate the value of momentum as:
$P \approx \dfrac{h}{{4\pi \Delta x}}$
So now the de broglie wavelength will be the following on substituting the momentum we get :
$\lambda = \dfrac{h}{{\dfrac{h}{{4\pi \Delta x}}}}$
$\lambda \approx 4\pi \Delta x$
Here we see that the de broglie wavelength is more than the size of the nucleus ($\Delta x$)

So the correct option is option B.

Note:The heisenberg uncertainty principle says that it is not possible for determining the position and velocity of the particle simultaneously. This principle arises from the duality of the wave. Each particle is associated with a wave as each particle has a wave-like behaviour.