Chapter 30, Problem 9P

To determine

If an alpha particle were released from rest near the surface of a ${}_{100}{}^{257}Fm$ nucleus, what would its kinetic energy be when far away?

Answer to Problem 9P

Solution:

The kinetic energy of alpha particle when is far away from ${}_{100}{}^{257}Fm$ is 30 MeV.

Explanation of Solution

Given info:

Initially, the alpha particle was released from rest ($K_i$= 0). Then, alpha particle is far away ($U_f=0$)from ${}_{100}{}^{257}Fm$ nucleus. The atomic number ($Z_a$)of alpha particle is 2, and the mass number ($A_a$)of alpha particle is 4. For ${}_{100}{}^{257}Fm$ nucleus its atomic number ($Z_F$)is 100 and its mass number ($A_F$)is 257.

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:

According to given info values:

$K_i=0$ (2)

And

$U_f=0$ (3)

Now, the initial potential energy ($U_i$)is:

$U_i=k\frac{q_a.q_F}{r_{aF}}$ (4)

Inserting (2), (3)and (4)into equation (1)we have:

$K_f=k\frac{q_a.q_F}{r_{aF}}$ (5)

Where:

• $q_a$ is total charge of alpha particle, which can be obtained using:

$q_a=Z_a.e=(1.6x{10}^{-19}C)Z_a$ (6)

Similarly, the charge for ${}_{100}{}^{257}Fm$ nucleus can be obtained using:

$q_F=Z_F.e=(1.6x{10}^{-19}C)Z_F$ (7)

• $r_{eF}$ is the distance between alpha particle and ${}_{100}{}^{257}Fm$ nucleus, which can be obtained using:

$r_{eF}=r_e+r_F=r_o{\left(A_e\right)}^{1/3}+r_o{\left(A_F\right)}^{1/3}$ (8)

Now, inserting (6), (7)and (8)into equation (5)we have:

$K_f=\frac{{\left(1.6\times {10}^{-19}C\right)}^2{Z}_e{Z}_F}{r_o\left(A_e^{1/3}+A_F^{1/3}\right)}$ (9)

Where:

$r_o=1.2\times {10}^{-15}m$ (10)