Question 45: Atoms & Nuclei

Question

Find the binding energy of the tritium nucleus:
[Given: mass of $$1 \mathrm{H}^3=3.01605 \mathrm{~u} ; \mathrm{~m}_{\mathrm{p}}=1.00782 \mathrm{~u} ; \mathrm{~m}_{\mathrm{n}}=1.00866 \mathrm{~u}$$.]

Accepted Answer

The binding energy of a nucleus is the energy required to disassemble the nucleus into its constituent protons and neutrons. The formula to calculate the binding energy is given by:
$$ 	ext{Binding Energy} = \left(	ext{Total mass of individual nucleons} - 	ext{Mass of the nucleus}ight) 	imes 931.5 \, 	ext{MeV/u} $$
For tritium ($$^3_1	ext{H}$$), it consists of 1 proton and 2 neutrons. We are given the following masses:
Mass of tritium nucleus, $$	ext{m}_{	ext{T}} = 3.01605 \, 	ext{u}$$
Mass of a proton, $$	ext{m}_{	ext{p}} = 1.00782 \, 	ext{u}$$
Mass of a neutron, $$	ext{m}_{	ext{n}} = 1.00866 \, 	ext{u}$$
First, we calculate the total mass of individual nucleons:
$$ 	ext{Total mass of nucleons} = 1 	imes 1.00782 \, 	ext{u} + 2 	imes 1.00866 \, 	ext{u}  $$
$$ = 1.00782 \, 	ext{u} + 2.01732 \, 	ext{u} $$
$$ = 3.02514 \, 	ext{u} $$
Next, we calculate the mass defect (difference between the total mass of individual nucleons and the mass of the nucleus):
$$ 	ext{Mass defect} = 	ext{Total mass of nucleons} - 	ext{Mass of the nucleus} $$
$$ = 3.02514 \, 	ext{u} - 3.01605 \, 	ext{u} $$
$$ = 0.00909 \, 	ext{u} $$
Now, we convert the mass defect into energy using the conversion factor 931.5 MeV/u:
$$ 	ext{Binding Energy} = 0.00909 \, 	ext{u} 	imes 931.5 \, 	ext{MeV/u} $$
$$ \approx 8.5 \, 	ext{MeV} $$
Therefore, the binding energy of the tritium nucleus is approximately
8.5 MeV
. Hence, the correct option is:
Option A: 8.5 MeV

Answer

8.5 MeV

Answer

8.5 J

Answer

0.00909 MeV

Answer

0.00909 eV

Atoms & Nuclei

Find the binding energy of the tritium nucleus: [Given: mass of $1 \mathrm{H}^3=3.01605 \mathrm{~u} ; \mathrm{~m}_{\mathrm{p}}=1.00782 \mathrm{~u} ; \mathrm{~m}_{\mathrm{n}}=1.00866 \mathrm{~u}$.]

Choose the correct answer: