Hint: Use Gauss law to find the electric field at the surface.

(a) Step 1: Find the electric field at the surface.

Let us suppose that universe is a perfect sphere of radius R and its constituent hydrogen atoms are distributed uniformly in the sphere.

As a hydrogen atom contains one proton and one electron, charge on each hydrogen atom. 

${\mathrm{e}}_{\mathrm{H}}={\mathrm{e}}_{\mathrm{P}}+\mathrm{e}=-(1+\mathrm{y})\mathrm{e}+\mathrm{e}=-\mathrm{ye}=\left(\mathrm{ye}\right)$

If E is electric field intensity at distance R, on the surface of the sphere, then according to Gauss' theorem,

$\oint \mathrm{E}.\mathrm{ds}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}$ i.e.,

 $\mathrm{E}\left(4{\mathrm{\pi R}}^2\right)=\frac{4}{3}\frac{{\mathrm{\pi R}}^3\mathrm{N}\left|\mathrm{ye}\right|}{{\mathrm{\epsilon }}_0}$
$\mathrm{E}=\frac{1}{3}\frac{\mathrm{N}\left|\mathrm{ye}\right|\mathrm{R}}{{\mathrm{\epsilon }}_0} ..............\left(\mathrm{i}\right)$

Step 2: Find the gravitational force on an atom at the surface.

Now, suppose, the mass of hydrogen atom$\simeq {\mathrm{m}}_{\mathrm{P}}=$Mass of a proton, G_R=gravitational field at distance R on the sphere.

Then, $-4{\mathrm{\pi R}}^2{\mathrm{G}}_{\mathrm{R}}=4\mathrm{\pi G} {\mathrm{m}}_{\mathrm{P}} \left(\frac{4}{3}{\mathrm{\pi R}}^3\right)\mathrm{N}$

$\Rightarrow {\mathrm{G}}_{\mathrm{R}}=\frac{-4}{3}{\mathrm{\pi Gm}}_{\mathrm{P}}\mathrm{NR} ..............\left(\mathrm{ii}\right)$

$\therefore$The gravitational force on this atom is ${\mathrm{F}}_{\mathrm{G}}={\mathrm{m}}_{\mathrm{P}}\times {\mathrm{G}}_{\mathrm{R}}=\frac{-4\mathrm{\pi }}{3}{\mathrm{Gm}}_{\mathrm{P}}^2\mathrm{NR} .............\left(\mathrm{iii}\right)$

Coulomb force on hydrogen atom at R is ${\mathrm{F}}_{\mathrm{C}}=\left(\mathrm{ye}\right) \mathrm{E}=\frac{1}{3}\frac{{\mathrm{Ny}}^2{\mathrm{e}}^2\mathrm{R}}{{\mathrm{\epsilon }}_0} [\mathrm{From} \mathrm{Eq}. (\mathrm{i}\left)\right]$

Step 3: Find the value of y.

Now, to start expansion ${\mathrm{F}}_{\mathrm{C}}>{\mathrm{F}}_{\mathrm{G}}$ and critical value of Y to start expansion would be when

${\mathrm{F}}_{\mathrm{C}}={\mathrm{F}}_{\mathrm{G}}$
$\Rightarrow \frac{1}{3}\frac{{\mathrm{Ny}}^2{\mathrm{e}}^2\mathrm{R}}{{\mathrm{\epsilon }}_0}=\frac{4\mathrm{\pi }}{3}{\mathrm{Gm}}_{\mathrm{P}}^2\mathrm{NR}$
$\Rightarrow {\mathrm{y}}^2=\left(4{\mathrm{\pi \epsilon }}_0\right)\mathrm{G}{\left(\frac{{\mathrm{m}}_{\mathrm{P}}}{\mathrm{e}}\right)}^2$
$=\frac{1}{9\times {10}^9}\times (6.67\times {10}^{-11})\left(\frac{{(1.66\times {10}^{-27})}^2}{{(1.6\times {10}^{-19})}^2}\right)=79.8\times {10}^{-38}$
$\Rightarrow \mathrm{y}=\sqrt{79.8\times {10}^{-38}}=8.9\times {10}^{-19}\simeq {10}^{-18}$

Thus, 10^-18 is the required critical value of Y corresponding to which expansion of the universe would start.

(b) Step 4: Find the net force on the atom at the surface.

Net force experience by the hydrogen atom is given by:

$\mathrm{F}={\mathrm{F}}_{\mathrm{C}}-{\mathrm{F}}_{\mathrm{G}}=\frac{1}{3}\frac{{\mathrm{Ny}}^2{\mathrm{e}}^2\mathrm{R}}{{\mathrm{\epsilon }}_0}-\frac{4\mathrm{\pi }}{3}{\mathrm{Gm}}_{\mathrm{p}}^2\mathrm{NR}$

Step 5: Find the net acceleration of the atom.

If the acceleration of hydrogen atom is represented by $\frac{{\mathrm{d}}^2\mathrm{R}}{{\mathrm{dt}}^2}$, then

${\mathrm{m}}_{\mathrm{P}}\frac{{\mathrm{d}}^2\mathrm{R}}{{\mathrm{dt}}^2}=\mathrm{F}=\frac{1}{3}\frac{{\mathrm{Ny}}^2{\mathrm{e}}^2\mathrm{R}}{{\mathrm{\epsilon }}_0}-\frac{4\mathrm{\pi }}{3}{\mathrm{Gm}}_{\mathrm{P}}^2\mathrm{NR}$
$=\left(\frac{1}{3}\frac{{\mathrm{Ny}}^2{\mathrm{e}}^2}{{\mathrm{\epsilon }}_0}-\frac{4\mathrm{\pi }}{3}{\mathrm{Gm}}_{\mathrm{P}}^2\mathrm{N}\right)\mathrm{R}$
$\therefore \frac{{\mathrm{d}}^2\mathrm{R}}{{\mathrm{dt}}^2}=\frac{1}{{\mathrm{m}}_{\mathrm{P}}}\left[\frac{1}{3}\frac{{\mathrm{Ny}}^2{\mathrm{e}}^2}{{\mathrm{\epsilon }}_0}-\frac{4\mathrm{\pi }}{3}{\mathrm{Gm}}_{\mathrm{P}}^2\mathrm{N}\right]\mathrm{R}={\mathrm{\alpha }}^2\mathrm{R} ..........\left(\mathrm{iv}\right)$

where, ${\mathrm{\alpha }}^2=\frac{1}{{\mathrm{m}}_{\mathrm{P}}}\left[\frac{1}{3}\frac{{\mathrm{Ny}}^2{\mathrm{e}}^2}{{\mathrm{\epsilon }}_0}-\frac{4\mathrm{\pi }}{3}{\mathrm{Gm}}_{\mathrm{P}}^2\mathrm{N}\right]$

Step 6: Find the velocity of expansion.

The general solution of Eq. (iv) is given by $\mathrm{R}={\mathrm{Ae}}^{\mathrm{\alpha t}}+{\mathrm{Be}}^{-\mathrm{\alpha t}}$. We are looking for expansion, here, so $\mathrm{B}=0$ and $\mathrm{R}={\mathrm{Ae}}^{\mathrm{\alpha t}}$.

$\Rightarrow$The velocity of expansion, $\mathrm{v}=\frac{\mathrm{dR}}{\mathrm{dt}}={\mathrm{Ae}}^{\mathrm{\alpha t}}\left(\mathrm{\alpha }\right)={\mathrm{\alpha Ae}}^{\mathrm{\alpha t}}=\mathrm{\alpha R}$

Hence, $\mathrm{v}\propto \mathrm{R}$ i.e., the velocity of expansion is proportional to the distance from the centre.