
The masses of neutrons and proton are 1.0087 a.m.u. and 1.0073 a.m.u. respectively. If the neutrons and protons combined to form a helium nucleus (alpha particles) of mass 4.0015 a.m.u., then the binding energy of the helium nucleus will be: (1 a.m.u. = 931 MeV)
A. 28.4 MeV
B. 20.8 MeV
C. 27.3 MeV
D. 14.2 MeV
Answer
588.6k+ views
Hint: Binding energy is the energy required to separate an atomic nucleus completely into its constituent subatomic particles i.e., protons and neutrons, or equivalently, the energy that would be liberated by combining individual protons and neutrons into a single nucleus. Mathematically, it is calculated as, \[B.E.=\Delta m{{c}^{2}}\]
Complete step by step answer:
We can find out the binding energy by applying the formula, \[B.E.=\Delta m{{c}^{2}}\]
Where,
\[\Delta m\] = mass defect
\[c\] = speed of light
Here, it is given that
Mass of neutron = 1.0087 a.m.u.
Mass of proton = 1.0073 a.m.u.
Mass of helium nucleus = 4.0015 a.m.u.
As we know that, in a helium nucleus there are 2 neutrons and 2 protons.
So theoretical mass will be,
\[=\left( 2\times 1.0087+2\times 1.0073 \right)\]
\[=4.032\text{ }a.m.u.\]
Mass defect \[\left( \Delta m \right)\] given by,
\[\Delta m=\left( 4.032-4.0015 \right)\text{ }a.m.u.\]
\[\Delta m=0.0305\text{ }a.m.u.\]
Now, by applying the formula, \[B.E.=\Delta m{{c}^{2}}\]
\[B.E.=\left( 0.0305\text{ }a.m.u. \right){{c}^{2}}\]
Since, \[1\text{ }a.m.u.\text{ }=931\text{ }MeV/{{c}^{2}}\]
Therefore,
\[B.E.=\left( 0.0305 \right)\times 931MeV\]
\[B.E.=28.4MeV\]
Hence, the correct option is A, i.e., 28.4 MeV
Additional Information:
This is true for all nuclei, that the mass of the nucleus is a little less than the mass of the individual neutrons and protons. This missing mass is known as the mass defect, and represents the binding energy of the nucleus.
For finding the binding energy, add the masses of the individual protons, neutrons, and electrons, subtract the mass of the atom, and convert that mass difference to energy.
Note: Firstly, students should understand the concept of binding energy and thereafter they need to memorize the famous formula for calculating it. Students should keep in mind that the value of 1 a.m.u. in units of \[MeV/{{c}^{2}}\], is \[931\text{ }MeV/{{c}^{2}}\] so that by putting this value, students can answer this type of questions in relatively less time.
Complete step by step answer:
We can find out the binding energy by applying the formula, \[B.E.=\Delta m{{c}^{2}}\]
Where,
\[\Delta m\] = mass defect
\[c\] = speed of light
Here, it is given that
Mass of neutron = 1.0087 a.m.u.
Mass of proton = 1.0073 a.m.u.
Mass of helium nucleus = 4.0015 a.m.u.
As we know that, in a helium nucleus there are 2 neutrons and 2 protons.
So theoretical mass will be,
\[=\left( 2\times 1.0087+2\times 1.0073 \right)\]
\[=4.032\text{ }a.m.u.\]
Mass defect \[\left( \Delta m \right)\] given by,
\[\Delta m=\left( 4.032-4.0015 \right)\text{ }a.m.u.\]
\[\Delta m=0.0305\text{ }a.m.u.\]
Now, by applying the formula, \[B.E.=\Delta m{{c}^{2}}\]
\[B.E.=\left( 0.0305\text{ }a.m.u. \right){{c}^{2}}\]
Since, \[1\text{ }a.m.u.\text{ }=931\text{ }MeV/{{c}^{2}}\]
Therefore,
\[B.E.=\left( 0.0305 \right)\times 931MeV\]
\[B.E.=28.4MeV\]
Hence, the correct option is A, i.e., 28.4 MeV
Additional Information:
This is true for all nuclei, that the mass of the nucleus is a little less than the mass of the individual neutrons and protons. This missing mass is known as the mass defect, and represents the binding energy of the nucleus.
For finding the binding energy, add the masses of the individual protons, neutrons, and electrons, subtract the mass of the atom, and convert that mass difference to energy.
Note: Firstly, students should understand the concept of binding energy and thereafter they need to memorize the famous formula for calculating it. Students should keep in mind that the value of 1 a.m.u. in units of \[MeV/{{c}^{2}}\], is \[931\text{ }MeV/{{c}^{2}}\] so that by putting this value, students can answer this type of questions in relatively less time.
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