Question

Find the Q-value and the kinetic energy of the $ \alpha $ particle, in the $ \alpha $ decay of
(a) $ {}_{88}^{226}Ra $ (b) $ _{86}^{222}Rn $
Given:
- $ m\left( {{}_{88}^{226}Ra} \right) = 226.02540u, $
- $ m\left( {{}_{86}^{222}Rn} \right) = 222.01750u, $
- $ m\left( {{}_{84}^{216}Po} \right) = 216.00189u $

Answer 1

Hint : When the masses are given in an atomic mass unit, the energy can be gotten by multiplying by 932 MeV. Mass of the alpha particle in the atomic mass unit is given as $ 4.00260u $ .
Formula used: In this solution we will be using the following formula;
 $ Q = \left[ {{M_r} - ({M_p} + {M_\alpha })} \right]{c^2} $ where $ Q $ is the Q-value, $ {M_r} $ is the mass of reactant, $ {M_p} $ is the mass of product, $ {M_\alpha } $ is mass of alpha particle and $ c $ is the speed of light.
 $ K{E_\alpha } = \dfrac{{A - 4}}{A}Q $ , where $ K{E_\alpha } $ is the kinetic energy of alpha particle, $ A $ is the mass number and $ Q $ is the Q value.

Complete step by step answer
In general, the Q value during an alpha decay is given as
 $ Q = \left[ {{M_r} - ({M_p} + {M_\alpha })} \right]{c^2} $
For the first case, the reaction is
 $ {}_{88}^{226}Ra \to _{86}^{222}Rn + {}_2^4He $ , hence Q value is given as,
 $ Q = \left[ {m\left( {{}_{88}^{226}Ra} \right) - m\left( {_{86}^{222}Rn} \right) - m\left( {{}_2^4He} \right)} \right]{c^2} $
 $ Q = \left( {226.02540u - 222.01750u - 4.00260u} \right){c^2} $ (since mass of alpha particle is $ 4.00260u $ )
By computation we get,
 $ Q = 0.0053u{c^2} = 0.0053 \times 931MeV $
 $ \Rightarrow Q = 4.9343MeV $
Now, the kinetic energy of alpha particle is given by
 $ K{E_\alpha } = \dfrac{{A - 4}}{A}Q $ , where $ K{E_\alpha } $ is the kinetic energy of alpha particle, $ A $ is the mass number of the reactant and $ Q $ is the Q value.
Hence,
 $ K{E_\alpha } = \dfrac{{226 - 4}}{{226}}\left( {4.9343MeV} \right) = \dfrac{{222}}{{226}}\left( {4.9343MeV} \right) $
By computation,
 $ K{E_\alpha } = 4.8470MeV $
Similarly the second reaction given by
 $ _{86}^{220}Rn \to {}_{84}^{216}Po + {}_2^4He $ ,
The Q value is given as
 $ Q = \left[ {m\left( {{}_{88}^{220}Rn} \right) - m\left( {_{86}^{216}Po} \right) - m\left( {{}_2^4He} \right)} \right]{c^2} $
Hence
 $ Q = \left( {220.01137u - 216.00189u - 4.00260u} \right){c^2} $
 $ \Rightarrow Q = 0.00688u{c^2} = 0.00688\left( {931MeV} \right) $
Hence, by computation we get,
 $ Q = 6.40528MeV $
Also,
 $ K{E_\alpha } = \dfrac{{220 - 4}}{{220}}\left( {6.40528MeV} \right) = \dfrac{{216}}{{226}}\left( {6.40528MeV} \right) $
Hence,
 $ K{E_\alpha } = 6.12186MeV $.

Note
As observed, the kinetic energy is always less than the Q value of any given nuclear reaction. This is because some of the Q values are directed back into the parent nucleus. This energy given to the parent nucleus is called recoil energy. This is true for not only alpha decay but beta as well. In beta decay not only the recoil energy and beta particles carry away energy but also a neutrino which is always emitted after beta decay.