Nuclear Physics: Question 5

Worked solution

Step 1: Find the total binding energy of the original nucleus X

$\text{BE}_X=236\times7.6=1793.6\text{ MeV}$

Step 2: Find the total binding energy of the two fragment nuclei Y

Each fragment Y has $118$ nucleons and $8.5\text{ MeV}$ of binding energy per nucleon, so one fragment has: $\text{BE}_Y=118\times8.5=1003\text{ MeV}$

There are two identical fragments produced, so their combined binding energy is: $\text{BE}_{2Y}=2\times1003=2006\text{ MeV}$

Step 3: Find the energy released

The energy released equals the increase in total binding energy, since more tightly-bound nucleons in the products correspond to a decrease in total mass: $E_{released}=\text{BE}_{2Y}-\text{BE}_X=2006-1793.6=212.4\text{ MeV}$

Recompute as a check, working per-fragment first: each fragment gains $8.5-7.6=0.9\text{ MeV}$ of binding energy per nucleon relative to X, over $118$ nucleons, giving $118\times0.9=106.2\text{ MeV}$ per fragment; doubling for both fragments gives $2\times106.2=212.4\text{ MeV}$, the same result.

Rounding to 3 significant figures, $E_{released}\approx\boxed{212}\text{ MeV}$.

Why the other options are wrong

• A ($106\text{ MeV}$): this is the energy gain of only one fragment relative to half of X’s binding energy, it omits the identical contribution from the second fragment.
• C ($1000\text{ MeV}$): this is just one fragment’s total binding energy ($1003\text{ MeV}$) quoted on its own, without subtracting the binding energy of the original nucleus X at all.
• D ($2010\text{ MeV}$): this is the combined binding energy of both fragments ($2006\text{ MeV}$) quoted directly, again without subtracting $\text{BE}_X=1793.6\text{ MeV}$ to find the actual increase.

Final answer

• The total energy released is $\boxed{212}\text{ MeV}$, option B.