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

Concept:

• The energy released in a nuclear reaction is given by \[ Q=\Delta mc^2 \]

• Using atomic mass units, \[ Q(\text{MeV}) = \Delta m \times 931.5 \] where \(\Delta m\) is in atomic mass units.

Step 1: Calculate the total mass of products.
\[ m_{\text{products}} = 234.043+4.003 \] \[ = 238.046\,u \]

Step 2: Find the mass defect.
\[ \Delta m = m_{\text{reactants}} - m_{\text{products}} \] \[ = 238.050-238.046 \] \[ = 0.004\,u \]

Step 3: Calculate the Q-value in MeV.
\[ Q = 0.004\times931.5 \] \[ = 3.726\ \text{MeV} \]

Step 4: Convert MeV into keV.
\[ 1\text{ MeV} = 1000\text{ keV} \] Therefore, \[ Q = 3.726\times1000 \] \[ = 3726\text{ keV} \] Hence, \[ \boxed{ Q=3726\text{ keV} } \] \[ \boxed{\text{Option (B)}} \]