Question 159: Atoms & Nuclei

Question

If $$M_0$$ is the mass of isotope $${ }_5^{12} B, M_p$$ and $$M_n$$ are the masses of proton and neutron, then nuclear binding energy of isotope is:

Accepted Answer

To determine the nuclear binding energy of the isotope $$_5^{12}B$$, we need to consider the mass defect concept. The mass defect is the difference between the sum of the individual masses of nucleons (protons and neutrons) and the actual mass of the nucleus.
Let's calculate the mass defect first. The isotope $$_5^{12}B$$ has 5 protons and 7 neutrons (since the total number of nucleons is 12). Therefore, the mass defect can be written as:
$$\Delta M = (5 M_p + 7 M_n) - M_0$$
Once we have the mass defect, the binding energy can be found using Einstein's mass-energy equivalence principle, which is given by:
$$E = \Delta M \cdot c^2$$
Substituting the mass defect into this equation, we get:
$$E = ((5 M_p + 7 M_n) - M_0) \cdot c^2$$
Therefore, the correct answer is Option A:
$$ (5 M_p + 7 M_n - M_0) \cdot c^2 $$

Answer

$$(5 M_p+7 M_n-M_o) C^2$$

Answer

$$(M_o-5 M_p-7 M_n) C^2$$

Answer

$$(M_o-5 M_p) C^2$$

Answer

$$(M_0-12 M_n) C^2$$

Atoms & Nuclei

If $M_0 is the mass of isotope { }_5^{12} B, M_p and M_n$ are the masses of proton and neutron, then nuclear binding energy of isotope is:

Choose the correct answer: