Chapter 33, Problem 15P

To determine

Q -value for the He burning reactions.

Answer to Problem 15P

Q -value for the He burning reactions producing Be is $-0.092\text{ MeV}$ and C product is $\text{7}\text{.366 MeV}$ .

Explanation of Solution

Given:

The mass of the unstable ${}_4^8\text{Be}$ is $8.005305\text{ u}$

Formula used:

Q -value is calculated as the difference between the mass of the reactants and the mass of the products.

Calculation:

Q -value for the reaction producing Be is,

  $\begin{array}{l}{}_2^4{\text{He + }}_2^4\text{He }\to {}_4^8\text{Be}\\ Q=m_{Be}{c}^2-2m_{He}{c}^2\\ Q=[2(4.002603\text{ u})-8.005305\text{ u}]{\text{c}}^2(931.5{\text{ MeV/c}}^2)\\ Q\text{ = }-\text{0}\text{.092 Mev}\end{array}$

Q-value for the reaction producing C is,

  $\begin{array}{l}{}_2^4{\text{He + }}_4^8\text{Be }\to {}_6^{12}\text{C}\\ Q=m_{Be}{c}^2+m_{He}{c}^2-m_C{c}^2\\ Q=[4.002603\text{ u + }8.005305\text{ u}-12.000000\text{ u}]{\text{c}}^2(931.5{\text{ MeV/c}}^2)\\ Q\text{ = 7}\text{.366 Mev}\end{array}$

Conclusion:

Thus, Q -value for the He burning reactions producing Be is $-0.092\text{ MeV}$ and C product is $\text{7}\text{.366 MeV}$ .