Question

In the study of Geiger-Marsdon experiment on scattering of $\alpha$-particles by a thin foil of gold, draw the trajectory of $\alpha$-particles in the coulomb field of target nucleus. Explain briefly how one gets the information on the size of the nucleus from this study.
From the relation $R=R_0{A}^{1/3}$, where $R_0$ is constant and A is the mass number of the nucleus, show that nuclear matter density is independent of A.
OR
Distinguish between nuclear fission and fusion. Show how in both these processes energy is released.
Calculate the energy release in MeV in the deuterium-tritium fusion reaction:
${}_1^{2}H{+}_1^{3}H{⟶}_2^{4}He+n$
Using the data:
$m\left({}_1^{2}H\right)=2.014102u$
$m\left({}_1^{3}H\right)=3.016049u$
$m\left({}_2^{4}He\right)=4.002603u$
$m_n=1.008665u$
$1u=931.5MeV/c^2$

Answer 1

Consider an alpha particle with initial $K.E=\frac{1}{2}m{v}^2$ directed towards the center of nucleus of an atom.

The force that exists between nucleus and $\alpha$ particle is coulomb's repulsive force. On account of this force, at the distance of closest approach $r_0$, the particle stops and cannot go further closer to the nucleus. And K.E gets converted to P.E

That is,

$\frac{1}{2}m{v}^2=\frac{Ze\left(2e\right)}{4\pi {ϵ}_0{r}_0}$

$⟹$ $r_0=\frac{Ze\left(2e\right)}{4\pi {ϵ}_0\left(\frac{1}{2}m{v}^2\right)}$

Hence, we can see that the size of the nucleus is approximately equal to the distance of the closest approach, $r_0$.

Now,

If ${}^{\prime }{m}^{\prime }$ is the average mass of the nucleon and ${}^{\prime }{r}^{\prime }$,the radius of the nucleus, then

Mass of the nucleus$=mA$ ; where $A$ is the mass number of the element

Volume of the nucleus, $v=\frac{4}{3}\pi r^3$

This implies,

$v=\frac{4}{3}\pi (R_0{A}^{\frac{1}{3}}{)}^3$

$⟹$ $v=\frac{4}{3}\pi R_0^{3}A$

$Density$ $of$ $nuclear$ $matter= Mass of nucleus /Volume of nucleus

That is,

$\rho =\frac{mA}{\frac{4}{3}\pi R_0^{3}A}$

$⟹$ $\rho =$$\frac{3m}{4\pi R_0^{3}A}$

Therefore, the nuclear density id independent of mass number A.