Atomic mass of `._6^13C` is 13.00335 amu and its mass number is 13.0. If 1 amu = 931 MeV. binding energy of the neutrons present in the nucleus is

Options

A. 0.24 MeV
B. 1.44 MeV
C. 1.68 MeV
D. 3.12 MeV

Answer

To solve the problem, we need to find the binding energy of the neutrons present in the nucleus of Carbon-13 (C-13). The steps to arrive at the solution are as follows:

### Step 1: Identify the Given Data
- Atomic mass of C-13: \( 13.00335 \, \text{amu} \)
- Mass number of C-13: \( 13 \)
- Conversion factor: \( 1 \, \text{amu} = 931 \, \text{MeV} \)

### Step 2: Calculate the Mass Defect
The mass defect (\( \Delta m \)) is calculated as:
\[ \Delta m = \text{Atomic mass} - \text{Mass number} \]
Substituting the values:
\[ \Delta m = 13.00335 \, \text{amu} - 13.0 \, \text{amu} = 0.00335 \, \text{amu} \]

### Step 3: Calculate the Binding Energy
The binding energy (\( BE \)) can be calculated using the mass defect:
\[ BE = \Delta m \times c^2 \]
Since \( 1 \, \text{amu} = 931 \, \text{MeV} \), we can express the binding energy in MeV as:
\[ BE = \Delta m \times 931 \, \text{MeV} \]
Substituting the mass defect:
\[ BE = 0.00335 \, \text{amu} \times 931 \, \text{MeV} = 3.12 \, \text{MeV} \]

### Step 4: Calculate the Number of Neutrons
For Carbon-13, the number of neutrons (\( N \)) can be calculated as:
\[ N = \text{Mass number} - \text{Atomic number} \]
The atomic number of Carbon is 6, so:
\[ N = 13 - 6 = 7 \]

### Step 5: Calculate the Binding Energy per Neutron
To find the binding energy specifically for the neutrons, we need to divide the total binding energy by the number of nucleons and then multiply by the number of neutrons:
\[ BE_{\text{per nucleon}} = \frac{BE}{\text{Mass number}} = \frac{3.12 \, \text{MeV}}{13} \]
Calculating this gives:
\[ BE_{\text{per nucleon}} \approx 0.24 \, \text{MeV} \]
Now, the total binding energy of the neutrons is:
\[ BE_{\text{neutrons}} = BE_{\text{per nucleon}} \times N = 0.24 \, \text{MeV} \times 7 = 1.68 \, \text{MeV} \]

### Final Answer
The binding energy of the neutrons present in the nucleus of Carbon-13 is approximately:
\[ \boxed{1.68 \, \text{MeV}} \]