It is experimentally found out that the mass of any nucleus is always less than the sum of the mass of its individual constituents. For example, consider the carbon-12 nucleus which is made up of 6 protons and 6 neutrons.

**Mass defect
and binding energy**

It is experimentally found out that
the mass of any nucleus is always less than the sum of the mass of its
individual constituents. For example, consider the carbon-12 nucleus which is
made up of 6 protons and 6 neutrons.

Mass of 6 neutrons = 6Ã—1.00866 *u *= 6.05196*u *

Mass of 6 protons = 6 Ã—1.00727 *u *= 6.04362 *u*

Mass of 6 electrons = 6 Ã— 0.00055*u *= 0.0033*u*

The
expected mass of carbon-12 nucleus

= 6.05196 *u *+ 6.04362 *u *= 12.09558*u*

But using mass spectroscopy, the atomic mass of carbon-12 atom is
found to be 12 *u*. So if we subtract
the mass of 6 electrons (0.0033 *u*)
from 12 *u*, we get the carbon-12
nuclear mass which is equal to 11.9967 *u*.
Note that the experimental mass of carbon-12 nucleus is less than the total
mass of its individual constituents by âˆ†*m
*= 0.09888*u *. This difference in
mass âˆ†*m* is called mass defect. In
general, if M, m_{p} and m_{n} are mass of the nucleus ( ^{A}_{z}*X *), the mass of a proton and the mass
of a neutron respectively, then the mass defect is given by

âˆ†*m *= (*Zm _{p} *+

Where has this mass disappeared? The answer was provided by
Albert Einstein with the help of famous mass-energy relation (*E *= *mc*^{2}
) . According to this relation, the mass can be converted into energy and
energy can be converted into mass. In the case of the carbon-12 nucleus, when 6
protons and 6 neutrons combine to form carbon-12 nucleus, mass equal to mass
defect disappears and the corresponding energy is released. This is called the
binding energy of the nucleus (BE) and is equal to (âˆ†*m*)*c*^{2} . In fact,
to separate the carbon-12 nucleus into individual constituents, we must supply
the energy equal to binding energy of the nucleus.

We can write the equation (8.20) in terms of binding energy

*BE *= (*Zm _{p} *+

It is always convenient to work with the mass of the atom than
the mass of the nucleus. Hence by adding and subtracting the mass of the *Z *electrons, we get

*BE *= (*Zm _{p} *+

*BE *= [*Z *(*m _{p} *+

where *m _{p} *+

*BE *= [*Zm _{H} *+

Here *M *+ *Zm _{e} *=

Finally, the binding energy in terms of the atomic masses is
given by

*BE *= [*Zm _{H} *+

Using Einsteinâ€™s mass-energy equivalence, the energy equivalent of
one atomic mass unit 1u = 1.66 Ã—10^{-27} Ã— (3 Ã—10^{8})^{2}

= 14.94 Ã—10^{-11}
J

â‰ˆ 931MeV

Compute the binding energy of ^{4}_{2}*He *nucleus using the following data:
Atomic mass of Helium atom, *M _{A}
*(

Binding energy *BE *= [*Zm _{H} *+

For helium nucleus, *Z *=
2, *N *= *Aâ€“Z *= 4â€“2 = 2

Mass defect

âˆ†*m *= [(2Ã—1.00785*u*) + (2 Ã—1.008665*u*) - 4.00260 *u*] âˆ†*m *= 0.03038*u*

*B*.*E *= 0.03038*u *Ã— *c*^{2}

*B*.*E *= 0.03038 Ã— 931*MeV *= 28 *MeV*

[1*uc*^{2} = 931*MeV *]

The binding energy of the
^{4}_{2}*He *nucleus
is 28 MeV.

Tags : Nuclear Physics , 12th Physics : UNIT 9 : Atomic and Nuclear Physics

Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail

12th Physics : UNIT 9 : Atomic and Nuclear Physics : Mass defect and binding energy | Nuclear Physics

**Related Topics **

Privacy Policy, Terms and Conditions, DMCA Policy and Compliant

Copyright Â© 2018-2023 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.