Question

Suppose a ${}_{88}^{226}Ra$ nucleus at rest and in ground state undergoes α-decay to a ${}_{86}^{222}Rn$ nucleus in its excited state. The kinetic energy of the emitted α particle is found to be 4.44 MeV. ${}_{86}^{222}Rn$ nucleus then goes to its ground state by γ-decay. The energy of the emitted γ photon is _______ keV.
[Given: atomic mass of ${}_{88}^{226}Ra$ = 226.005 u, atomic mass of ${}_{86}^{222}Rn$ = 222.000 u, atomic mass of α particle = 4.000 u, 1 u = 931 MeV/${c^2}$, c is speed of the light]

Answer 1

Hint: This question can be answered by using newton’s second law. According to Newton's second law for a particular system rate of change of momentum will be equal to the external force acting on the system. Here the decay process is considered as an internal action.

Formula used:
${F_{ext}} = \dfrac{{d{p_{system}}}}{{dt}}$
$K.E = \dfrac{{{p^2}}}{{2m}}$
$Q = \Delta m{c^2}$

Complete step by step answer:
Initially the radium nucleus is at rest and then it decayed into the alpha particle and radon nucleus. Masses of the radium, alpha particle and radon nuclei are given. First we will find out the difference in the reactants and product masses which is called mass defect. Later we will multiply it with ${c^2}$ value to get the total heat(Q) produced in that reaction. Out of that heat some energy is utilized as the kinetic energy of the products and the remaining energy is emitted in the gamma emission.
Initially let us find the mass defect
$\Delta m$= [226.005-222-4] = 0.005 amu
Then the value of Q will be $Q = \Delta m{c^2} = 0.005 \times 931.5 = 4.655MeV$
${F_{ext}} = \dfrac{{d{p_{system}}}}{{dt}}$
During this decay external force is zero as everything happens internally so we can conserve the momentum of the system
Initial momentum = 0 as radium is at rest.
Reactants momentum = products momentum
$ \Rightarrow $0 = ${p_\alpha } + {p_{Rn}}$
$ \Rightarrow {p_\alpha } = - {p_{Rn}}$
$ \Rightarrow K.E = \dfrac{{{p^2}}}{{2m}}$
When magnitudes of momentum are equal then kinetic energy is inversely proportional to mass
$\eqalign{
  & \dfrac{{K.{E_\alpha }}}{{K.{E_{Rn}}}} = \dfrac{{m{}_{Rn}}}{{{m_\alpha }}} \cr
  & \Rightarrow K.{E_{Rn}} = \dfrac{{{m_\alpha }}}{{m{}_{Rn}}}K.{E_\alpha } \cr
  & \Rightarrow K.{E_{Rn}} = \dfrac{4}{{222}} \times 4.44 = 0.08MeV \cr} $
So total kinetic energy of products will be $4.44 + 0.08 = 4.520$
Difference between heat released and kinetic energy of products is $4.655 - 4.520 = 0.135MeV = 135keV$

So energy of gamma photon would be 135 kilo electron volts

Note: It is a bit based on conservation of energy too because energy can neither be created nor destroyed and it is just converted from one form to another form. Here total heat produced is converted into the kinetic energy of products and energy of the photon produced during the gamma emission to the ground state.