Question

Consider the nuclear reaction \[ ^{238}\mathrm{U} \rightarrow ^{234}\mathrm{Th} + ^{4}\mathrm{He} \] Take masses of \(^{238}\mathrm{U}\), \(^{234}\mathrm{Th}\), and \(^{4}\mathrm{He}\) as \[ 238.050\,u,\qquad 234.043\,u,\qquad 4.003\,u \] respectively. The \(Q\)-value for the reaction, in keV, is: \[ 1u = 931.5\ \mathrm{MeV}/c^2 \]

• A. 3740
• B. 3726
• C. 3730
• D. 3736

Hint:

For nuclear reactions, \[ Q=(\text{Mass defect})\times931.5\ \text{MeV} \] Always calculate the mass defect first and then convert the energy into the required units.

Answer 1

The Correct Option is B

Step 1: Recall the Q-value idea.
The energy released equals the mass lost times $c^2$. Using atomic mass units, $Q(\text{MeV}) = \Delta m \times 931.5$, where $\Delta m$ is the mass defect in $u$.
Step 2: Add the product masses.
For the products: $234.043 + 4.003 = 238.046\,u$.
Step 3: Find the mass defect.
$\Delta m = 238.050 - 238.046 = 0.004\,u$.
Step 4: Convert to MeV.
$Q = 0.004 \times 931.5 = 3.726$ MeV.
Step 5: Convert to keV.
Since $1$ MeV $= 1000$ keV, $Q = 3.726 \times 1000 = 3726$ keV.
Step 6: Match the option.
This equals option B.
\[ \boxed{ Q = 3726 \text{ keV} } \]