Question 72: Atoms & Nuclei

Question

Assuming the experimental mass of ${ }_6^{12} C$ as $12 u$, the mass defect of ${ }_6^{12} C$ atom is $\_\_\_\_$ $\mathrm{MeV} / \mathrm{c}^2$.
(Mass of proton $=1.00727 \mathrm{u}$. mass of neutron $=1.00866 \mathrm{u}, 1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^2$ and c is the speed of the light in vacuum).

Accepted Answer

Mass defect $(\Delta \mathrm{m})$ is the difference between the sum of the masses of the individual nucleons (protons and neutrons) and the actual experimental mass of the nucleus.
For the atom $\mathrm{C}_6^{12}$, atomic number is $\mathrm{Z}=6$ (Number of protons)
Mass number is $\mathrm{A}=12$
Number of neutrons $\mathrm{N}=\mathrm{A}-\mathrm{Z}=12-6=6$
So, mass of 6 protons $=6 	imes 1.00727 u=6.04362 u$
And the mass of 6 neutrons $=6 	imes 1.00866 u=6.05196 u$
So, total theoretical mass $=6.04362+6.05196=12.09558 \mathrm{u}$
It is given that the experimental mass of $\mathrm{C}_6^{12}=12 \mathrm{u}$ :
Hence, the mass defect is
$$ \Delta \mathrm{m}=	ext { (Total theoretical mass) }- 	ext { (Experimental mass) } $$
$$\Rightarrow $$ $$ \Delta \mathrm{m}=12.09558 \mathrm{u}-12 \mathrm{u}=0.09558 \mathrm{u} $$
The energy equivalent of 1 u is given as $931.5 \mathrm{MeV} / \mathrm{c}^2$.
$$ \Delta \mathrm{m}\left(	ext { in } \mathrm{MeV} / \mathrm{c}^2ight)=0.09558 	imes 931.5 $$
$$\Rightarrow $$$$ \Delta \mathrm{m} \approx 89.03277 \approx 89.03 \mathrm{MeV} / \mathrm{c}^{\wedge} 2 $$
Therefore, the mass defect of the $\mathrm{C}_6^{12}$ atom is approximately $89.03 \mathrm{MeV} / \mathrm{c}^2$.
Thus, the correct option is (B).

Answer

127.5

Answer

89.03

Answer

272.0

Answer

92.0

Atoms & Nuclei

Assuming the experimental mass of { }_6^{12} C as 12 u, the mass defect of { }_6^{12} C atom is \_\_\_\_ \mathrm{MeV} / \mathrm{c}^2. (Mass of proton =1.00727 \mathrm{u}. mass of neutron =1.00866 \mathrm{u}, 1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^2 and c is the speed of the light in vacuum).

Choose the correct answer: