91Ó°ÊÓ

Given information:

Distance between the electrons and proton is $2.0nm$.

Distance between two electron points are $0.50nm$.

Find:

The objective is to find potential energy of the electron-proton interactions.

The momentum of the mechanical electron density, including the centrifugal acceleration through valence activity, contribute up the average system.

The elastic efficiency is defined utilizing $KE=1/2mv2$. This will be used to quantify the asteroid's together throughout.

We simply need to find the separation seen between radicals and the hydrogen since the possibilities is a multidimensional parameter.

Per the Mathematicians, the displacement will be

$d=\sqrt{2.0^2+0.5^2}=\underset{¯}{2.0616\mathrm{nm}}$

As a result, one electron's potential will be

$U_1=k\frac{-e^2}{d}=\frac{9×{10}^9-{\left(1.6×{10}^{-19}\right)}^2}{2.0616×{10}^{-9}}=-1.1×{10}^{-19}\mathrm{J}$

Therefore, $-2.2×{10}^{-19}\mathrm{J}$

We understand that the velocity of the quark is mostly because of asymmetry and the principle that quadratic constants require only be supplied multiple positions.

$U=-2.2×{10}^{19}\mathrm{J}$