Question 33: Atoms & Nuclei

Question

In a nuclear fusion reaction, two nuclei, $A$ and $B$ fuse to produce a nucleus $C$, releasing an amount of energy $\Delta \mathrm{E}$ in the process. If the mass defects of the three nuclei are $\Delta M_A, \Delta M_B$ and $\Delta M_C$ respectively, then which of the following relations is true? ( $c$ is the speed of light).

Accepted Answer

The key is that the binding energy of a nucleus is
$$B=\Delta M\,c^2$$
where ÃÂM is its massÃ¢ÂÂdefect.  In the fusion
$$A + B 	o C + 	ext{(energy } \Delta E	ext{)}$$
the energy released is the increase in binding energy, i.e.
$ \Delta E \;=\; B_C - (B_A + B_B)            \;=\; \bigl(\Delta M_C - (\Delta M_A + \Delta M_B)\bigr)\,c^2 $
Rearranging gives
$ \Delta M_A + \Delta M_B    = \Delta M_C \;-\;\frac{\Delta E}{c^2} $
which is Option D.

Answer

$\Delta M_A+\Delta M_B=\Delta M_C+\frac{\Delta E}{c^2}$

Answer

$\Delta M_A-\Delta M_B=\Delta M_C+\frac{\Delta E}{c^2}$

Answer

$\Delta M_A-\Delta M_B=\Delta M_C-\frac{\Delta E}{c^2}$

Answer

$\Delta M_A+\Delta M_B=\Delta M_C-\frac{\Delta E}{c^2}$

Atoms & Nuclei

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