Question

Consider an $\alpha$-particle just in contact with a ${}_{92}^{238}\text{U}$ nucleus. The Coulombic repulsion energy (i.e, the height of the Coulombic barrier between ${}^{238}\text{U}$ and alpha particle) assuming that the distance between them is equal to the sum of their radii is

• A. $16.35MeV$
• B. $46.66MeV$
• C. $22.24MeV$
• D. $26.14MeV$

Answer 1

The correct option is D $26.14MeV$

The expression for the radius of the nucleus is as shown below.
$r_{nucleus}=1.3\times {10}^{-13}(A{)}^{1/3}$; where $A$ is mass number
Radius of ${}_{92}^{238}\text{U}=1.3\times {10}^{-13}\times (238{)}^{1/3}$
$=8.06\times {10}^{-13}cm$
Radius of ${}_2^4\text{He}=1.3\times {10}^{-13}\times (4{)}^{1/3}$
$=2.06\times {10}^{-13}cm$
Total distance between uranium and helium nuclei is equal to the sum of their radii.

It is $=(8.06+2.06)\times {10}^{-13}=10.12\times {10}^{-13}cm$
The Coulombic repulsion energy is:

$\frac{Q_1{Q}_2}{r}$ $=\frac{92\times 4.8\times {10}^{-10}\times 2\times 4.8\times {10}^{-10}}{10.12\times {10}^{-13}}erg$ (because $Q_1$ and $Q_2$ in esu and r in cm)

$=418.9\times {10}^{-7}erg=418.9\times {10}^{-14}$J

$=418.9\times {10}^{-14}/1.602\times {10}^{-19}eV$

$=\frac{26.14\times {10}^6}{{10}^6}MeV$

$=26.14MeV$
Hence, the coulombic repulsion energy is $26.14MeV$.