Chapter 30, Problem 21P

To determine

Part (a):

The nucleus ${}_4{}^8\text{B}\text{e}$ is unstable and will decay into 2 $\alpha$ particles.

Answer to Problem 21P

Solution:

Since binding energy of ${}_4{}^8\text{B}\text{e}$ is lesser than that of two He nucleus, ${}_4{}^8\text{B}\text{e}$ will decay into two alpha particles.

Explanation of Solution

Formula used:

Binding energy is $\Delta m{c}^2$.

The binding energy of two Helium nuclei is $56.4\text{MeV}$. Since binding energy of ${}_4{}^8\text{B}\text{e}$ is lesser than that of two He nucleus, ${}_4{}^8\text{B}\text{e}$ will decay into two alpha particles.

Calculation:

Mass of ${}_4{}^8\text{B}\text{e}$ is $8.005305\text{u}$. Mass of neutron ${}_0{}^1\text{n}$ is $1.008665\text{u}$. Mass of proton ${}_1{}^1\text{H}$ is $1.007825\text{u}$.

Mass defect is $\Delta \text{m}=4\times 1.007825\text{u}+\times 41.008665\text{u}-8.005305\text{u}=\text{0}\text{.058459}\text{u}$.

Binding energy $\text{0}\text{.058459}\text{u}\times \frac{931.5\text{MeV}}{\text{u}}=54.455\text{MeV}$.

To determine

Part (b):

The nucleus ${}_6{}^{12}\text{C}$ is stable against decay into 3 $\alpha$ particles.

Answer to Problem 21P

Solution:

Yes. Binding energy of ${}_6{}^{12}\text{C}$ is positive.

Explanation of Solution

Formula used:

Binding energy is $\Delta m{c}^2$.

Binding energy of ${}_6{}^{12}\text{C}$ is positive. Hence, ${}_6{}^{12}\text{C}$ is stable against decay into 3 $\alpha$ particles.

Calculation:

Mass of ${}_6{}^{12}\text{C}$ is $12.000000\text{u}$. Mass of neutron ${}_0{}^1\text{n}$ is $1.008665\text{u}$. Mass of proton ${}_1{}^1\text{H}$ is $1.007825\text{u}$.

Mass defect is $\Delta \text{m}=6\times 1.007825\text{u}+6\times 1.008665\text{u}-12.0\text{u}=\text{0}\text{.09894}\text{u}$.

Binding energy $\text{0}\text{.09894}\text{u}\times \frac{931.5\text{MeV}}{\text{u}}=\text{92}\text{.16}\text{MeV}$.