NCERT Exemplar Class 12 Physics Solutions Chapter 1 Long Answer
Question:1.26
In 1959, Lyttleton and Bondi suggested that the expansion of the universe could be explained if matter carried a net charge. Suppose that the universe is made up of hydrogen atoms with a number density N, which is maintained a constant. Let the charge on the proton be: ep = – (1 + y)e where e is the electronic charge.
(a) Find the critical value of y such that expansion may start.
(b) Show that the velocity of expansion is proportional to the distance from the centre.
Answer:
(a) Let us suppose that universe is a perfect sphere of radius $R$ and its constituent hydrogen atoms are distributed uniformly in the sphere.
As hydrogen atom contains one proton and one electron, charge on each hydrogen atom.
$
e_H=e_P+e=-(1+Y) e+e=-Y e=(Y e)
$
If $E$ is electric field intensity at distance $R$, on the surface of the sphere, then according to Gauss' theorem,
$
\begin{aligned}
\oint \text { E.ds } & =\frac{q}{\varepsilon_0} \text { i.e., } E\left(4 \pi R^2\right)=\frac{4}{3} \frac{\pi R^3 N|Y e|}{\varepsilon_0} \\
E & =\frac{1}{3} \frac{N|Y e| R}{\varepsilon_0} --------(i)
\end{aligned}
$
Now, suppose, mass of each hydrogen atom $\simeq m_P=$ Mass of a proton, $G_R=$ gravitational field at distance $R$ on the sphere.
Then
$
\begin{aligned}
-4 \pi R^2 G_R & =4 \pi G m_P\left(\frac{4}{3} \pi R^3\right) N \\
G_R & =\frac{-4}{3} \pi G m_P N R ---------(ii)
\end{aligned}
$
$\therefore$ Gravitational force on this atom is:
$F_G=m_P \times G_R=\frac{-4 \pi}{3} G m_P^2 N R$-----------(iii)
Coulomb force on hydrogen atom at $R$ is:
$F_C=(Y e) E=\frac{1}{3} \frac{N Y^2 e^2 R}{\varepsilon_0}$ ------------[from Eq. (i)]
Now, to start expansion $F_C>F_G$ and critical value of $Y$ to start expansion would be when
$
\begin{array}{rlrl}
\Rightarrow& F_C =F_G \\
\Rightarrow & \frac{1}{3} \frac{N Y^2 e^2 R}{\varepsilon_0} =\frac{4 \pi}{3} G m_P^2 N R \\
\Rightarrow & Y^2 =\left(4 \pi \varepsilon_0\right) G\left(\frac{m_P}{e}\right)^2 \\
\Rightarrow & Y^2 =\frac{1}{9 \times 10^9} \times\left(6.67 \times 10^{-11}\right)\left(\frac{\left(1.66 \times 10^{-27}\right)^2}{\left(1.6 \times 10^{-19}\right)^2}\right)\\
\Rightarrow & Y^2=79.8 \times 10^{-38} \\
\Rightarrow & Y =\sqrt{79.8 \times 10^{-38}}=8.9 \times 10^{-19} \simeq 10^{-18}
\end{array}
$
Thus, $10^{-18}$ is the required critical value of $Y$ corresponding to which expansion of universe would start.
(b) Net force experience by the hydrogen atom is given by
$
F=F_C-F_G=\frac{1}{3} \frac{N Y^2 e^2 R}{\varepsilon_0}-\frac{4 \pi}{3} G m_P^2 N R
$
If acceleration of hydrogen atom is represent by $d^2 R / d t^2$, then
$
\begin{aligned}
& m_p \frac{d^2 R}{d t^2}=F=\frac{1}{3} \frac{N Y^2 e^2 R}{\varepsilon_0}-\frac{4 \pi}{3} G m_p^2 N R \\
& =\left(\frac{1}{3} \frac{N Y^2 e^2}{\varepsilon_0}-\frac{4 \pi}{3} G m_p^2 N\right) R \\
& \therefore \quad \frac{d^2 R}{d t^2}=\frac{1}{m_p}\left[\frac{1}{3} \frac{N Y^2 e^2}{\varepsilon_0}-\frac{4 \pi}{3} G m_p^2 N\right] R=\alpha^2 R \\
& \text { where, } \\
& \alpha^2=\frac{1}{m_p}\left[\frac{1}{3} \frac{N Y^2 e^2}{\varepsilon_0}-\frac{4 \pi}{3} G m_P^2 N\right]
\end{aligned}
$
The general solution of Eq. (iv) is given by $R=A e^{\alpha t}+B e^{-\alpha t}$.
We are looking for expansion, here, so $B=0$ and $R=A e^{\alpha t}$.
$\Rightarrow \quad$ Velocity of expansion, $v=\frac{d R}{d t}=A e^{\alpha t}(\alpha)=\alpha A e^{\alpha t}=\alpha R$
Hence, $v \propto R$ i.e., velocity of expansion is proportional to the distance from the centre.
