Question

If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is .......... × 10^–2 MeV.
(Given 1 u = 931 MeV/c² , atomic mass of helium = 4.002603 u)

Answer 1

Correct Answer: 727

Given reaction:

\[ ^3_2\text{He} \longrightarrow ^{12}_6\text{C} + \gamma \text{ rays} \]

Mass defect:

\[ \Delta m = (3m_{\text{He}} - m_{\text{C}}) \]

Calculating:

\[ \Delta m = (3 \times 4.002603 - 12) = 0.007809 \, \text{u} \]

Energy released:
\[ \text{Energy} = 931 \Delta m \, \text{MeV} \] \[ = 7.27 \, \text{MeV} = 727 \times 10^{-2} \, \text{MeV} \]

Answer 2

Step 1: Write down the given data
We are told that three helium nuclei (each with mass \(4.002603\,\text{u}\)) combine to form one carbon nucleus (\(^{12}\text{C}\)). The given constants are:
\[ 1\,\text{u} = 931\,\text{MeV}/c^2. \] Hence:
Mass of 3 helium atoms = \(3 \times 4.002603 = 12.007809\,\text{u}.\)
Mass of 1 carbon atom = \(12.000000\,\text{u}\) (by definition).

Step 2: Compute the mass defect
\[ \Delta m = \text{(mass of 3 He)} - \text{(mass of 1 C)} = 12.007809 - 12.000000 = 0.007809\,\text{u}. \]

Step 3: Convert mass defect to energy
\[ E = \Delta m \times 931 = 0.007809 \times 931 = 7.27\,\text{MeV}. \] This energy corresponds to the total energy released in the reaction:
\[ 3\,{}^{4}\text{He} \;\longrightarrow\; {}^{12}\text{C} + \text{energy}. \]

Step 4: Express in the given format
They ask for the value in the form “…… × 10⁻² MeV”. Since \(7.27 = 727 \times 10^{-2}\,\text{MeV}\), the required value is:
\[ \boxed{727}. \]

Final answer

727