Question

The binding energy of ${}_{17}C{l^{35}}$ nucleus is 298 MeV. Find its atomic mass. The mass of the hydrogen atom (${}_1{H^1}$) is 1.008143 a.m.u. Given 1 a.m.u. =931 MeV.

Answer 1

Hint: If we take an example of any atom, initially the sum of masses of protons and neutrons which are free will be more than their mass after forming into an atom. This difference in mass is known as mass defect and we will use it to find atomic mass of chlorine.
Formula used:
$B.E = \Delta m{c^2}$
${c^2} = 931\dfrac{{Mev}}{{a.m.u}}$

Complete answer:
Law of conservation of mass states that the mass can never be created nor destroyed, it is just converted from one form to the other. During the mass defect the entire mass which is lost should be appeared in some or the other manner. Here the mass will be appeared in the form of binding energy which will be denoted as B.E and its formula is given as
$B.E = \Delta m{c^2}$
Where
$\Delta m$ is the difference in masses called as mass defect
‘c’ is the velocity of light
In the given question mass of hydrogen atom is given i.e we can consider it as mass of proton.
So mass of proton(${m_p}$) is 1.008143 a.m.u
Mass of neutron is(${m_n}$) 1.008986 a.m.u
Chlorine atom has 17 protons and 18 neutrons in its nucleus.
Hence mass of 17 protons is
$17{m_p} = 17 \times 1.008143a.m.u = 17.138431a.m.u$
Mass of 18 neutrons is
$18{m_n} = 18 \times 1.008986 = 18.161748a.m.u$
Hence the total mass of 17 protons and 18 neutrons will be
$17{m_p} + 18{m_n} = 17.138431a.m.u + 18.161748a.m.u = 35.300179a.m.u$
The formula for binding energy is
$B.E = \Delta m{c^2}$, ${c^2} = 931\dfrac{{Mev}}{{a.m.u}}$
$\eqalign{
  & \Rightarrow \dfrac{{B.E}}{{{c^2}}} = \Delta m \cr
  & \Rightarrow \Delta m = \dfrac{{298}}{{931}} \cr
  & \Rightarrow \Delta m = 0.320085a.m.u \cr} $
Atomic mass will be less than the sum of individual nucleons by the mass defect amount. Let atomic mass be M
So
$\eqalign{
  & M = \left( {17{m_p} + 18{m_n}} \right) - \left( {\Delta m} \right) \cr
  & \Rightarrow M = 35.300179 - 0.320085 = 34.980094a.m.u \cr} $
Hence the mass of chlorine atom will be $34.980094a.m.u$

Note:
The difference in the mass will appear as the energy and that energy will be generally utilized by the products as the kinetic energy. So in nuclear reactions the heat energy released due to the mass defect appears as the kinetic energy of the products.