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Atoms & Nuclei

JEE Main level, single correct, medium

181 of 239

Question 181 of 239

Assuming the experimental mass of ${}^{12}_{6}C$ as $12\text{ u}$ , the mass defect of ${}^{12}_{6}C$ atom is _______ $\text{MeV}/c^2$ .
(Mass of proton $= 1.00727\text{ u}$ , mass of neutron $= 1.00866\text{ u}$ , $1\text{ u} = 931.5\text{ MeV}/c^2$ and $c$ is the speed of the light in vacuum).

Show answer and solution

Answer: (B) $89.03$

The mass of

6

protons is

6 \times 1.00727\text{ u} = 6.04362\text{ u}

.

The mass of

6

neutrons is

6 \times 1.00866\text{ u} = 6.05196\text{ u}

.

Total mass of the constituent nucleons is

6.04362\text{ u} + 6.05196\text{ u} = 12.09558\text{ u}

.

The mass defect

\Delta m

is the difference between the total mass of the nucleons and the experimental mass of the atom (ignoring electron mass as per the given data):

\Delta m = 12.09558\text{ u} - 12\text{ u} = 0.09558\text{ u}

Converting the mass defect into

\text{MeV}/c^2

\Delta m = 0.09558 \times 931.5\text{ MeV}/c^2 = 89.03277\text{ MeV}/c^2 \approx 89.03\text{ MeV}/c^2

Answer:

89.03

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