Question:

Assuming the experimental mass of \( {}^{12}_{6}\text{C} \) as 12 u, the mass defect of \( {}^{12}_{6}\text{C} \) atom is____MeV/\( c^2 \). 
(Mass of proton = 1.00727 u, mass of neutron = 1.00866 u, 1 u = 931.5 MeV/\( c^2 \))

Updated On: Apr 12, 2026
  • 127.5
  • 89.03
  • 272.0
  • 92.0
Show Solution
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
Mass defect (\( \Delta m \)) is the difference between the total mass of the constituent nucleons (protons and neutrons) and the actual experimental mass of the atom. Carbon-12 has 6 protons and 6 neutrons.
: Key Formula or Approach:
\( \Delta m = [Z \cdot m_p + (A-Z) \cdot m_n] - M_{\text{experimental}} \).
Energy equivalent \( E = \Delta m \cdot 931.5 \text{ MeV/u} \).
Step 2: Detailed Explanation:
For \( {}^{12}_{6}\text{C} \):
Number of protons \( Z = 6 \).
Number of neutrons \( A - Z = 12 - 6 = 6 \).
Total mass of nucleons:
\[ M_{\text{nucleons}} = 6 \times 1.00727 + 6 \times 1.00866 \]
\[ M_{\text{nucleons}} = 6.04362 + 6.05196 = 12.09558 \text{ u} \].
Experimental mass \( M_{\text{exp}} = 12.00000 \text{ u} \).
Mass defect:
\[ \Delta m = 12.09558 - 12.00000 = 0.09558 \text{ u} \].
Convert mass defect to MeV/\( c^2 \):
\[ \Delta E = 0.09558 \times 931.5 \text{ MeV/}c^2 \]
\[ \Delta E \approx 89.03277 \text{ MeV/}c^2 \].
Step 3: Final Answer:
The mass defect is 89.03 MeV/\( c^2 \).
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