Chapter 30, Problem 12P

To determine

How much kinetic energy, in MeV, must a $\alpha$ particle have to just “touch” the surface of a ${}_{92}{}^{232}U$ nucleus?

Answer to Problem 12P

Solution:

The kinetic energy for $\alpha$ particle that just “touch” the surface of a ${}_{92}{}^{232}U$ nucleus is 29 MeV.

Explanation of Solution

Given info:

The atomic number ($Z_a$)of alpha particle is 2 and the mass number ($A_a$)is 4. For ${}_{92}{}^{232}U$ nucleus its atomic number ($Z_U$)is 92 and themass number ($A_U$)is 232.

Formula used:

The conservation of total energy is defined as:

$U_i+K_i=U_f+K_f$ (1)

Where:

• $U_i$, $U_f$ are respectively the initial and final potential energy.

• $K_i$, $K_f$ are respectively the initial and final kinetic energy.

Calculation:

Assuming that ${}_{92}{}^{232}U$ nucleus is at rest, then; initially before impact alpha particle ($\alpha$)has a kinetic energy ($K_i$)and a value of zero for potential energy ($U_i$). Now, when the alpha particle “touch” the surface of ${}_{92}{}^{232}U$ nucleus its kinetic energy ($K_f$)is zero and potential energy ($U_f$)is:

$U_f=k\frac{q_a{q}_U}{r_{aU}}$ (2)

Where $q_a$, the total charge of alpha particle is:

$q_a=Z_{a}e=(1.6\times {10}^{-19}C)Z_a$ (3)

Similarly, $q_U$ is called total charge of ${}_{92}{}^{232}U$ nucleus which can be obtained using:

$q_U=Z_{U}e=(1.6\times {10}^{-19}C)Z_U$ (4)

Now, the distance $r_{aU}$ between centers of alpha particle and ${}_{92}{}^{232}U$ nucleus when alpha particle “touch” the nucleus is:

$r_{aU}=r_a+r_U=r_o\left(A_a^{1/3}+A_U^{1/3}\right)=(1.2\times {10}^{-15}m)\left(A_a^{1/3}+A_U^{1/3}\right)$ (5)

Now, inserting (3), (4)and (5)into equation (2)we have:

$U_f=k\frac{{(1.6\times {10}^{-19}C)}^2{Z}_a{Z}_U}{(1.2\times {10}^{-15}m)(A_a^{1/3}+A_U^{1/3})}$ (6)