Question

A nucleus of mass number 220 decays by \[\alpha \]decay. The energy released in the reaction is \[5MeV\]. The kinetic energy of an \[\alpha \]-particle is:
(A) \[\dfrac{1}{{54}}MeV\]
(B) \[\dfrac{{27}}{{11}}MeV\]
(C) \[\dfrac{{54}}{{11}}MeV\]
(D) \[\dfrac{{55}}{{54}}MeV\]

Answer 1

Hint Write the \[\alpha \]reaction of a particle with mass number 220, giving out 4 He and Y particles with mass number 216. Add the sum of kinetic energy to kinetic energy and it is said that the momentum after the decay is conserved.

Complete step by step Solution
Alpha decay is a type of decay of radioactive materials where an atomic nucleus with a specified mass emits an helium nucleus (which is also called as alpha particle) and hence transform the decayed particle into different atomic nucleus with a mass number reduced by a factor of 4 and atomic number reduced by a factor of 2. Applying this to our scenario we get,
\[{}_Z^{220}X \to {}_{Z - 2}^{216}Y + {}_2^4He\]
Now , it is said that the overall energy released by the reaction to be\[5MeV\]. Now , total energy is given as the sum of kinetic energy of each particle that is decayed. Hence, we can write the equation as
\[ \Rightarrow \dfrac{1}{2}{m_y}{v_y}^2 + \dfrac{1}{2}{m_{He}}{v_{He}}^2 = 5MeV\](Conservation of energy)
\[ \Rightarrow \dfrac{1}{2}(216){v_y}^2 + \dfrac{1}{2}(4){v_{He}}^2 = 5MeV\]-----(1)
Now, in this type of decay, it is understood that the momentum of the decayed particle is equal to that of momentum of the alpha particle produced, thus the linear momentum is conserved. According to law of conservation of momentum, we get
\[ \Rightarrow {m_y}{v_y} = {m_{He}}{v_{He}}\]
Taking the known quantities on one side we get,
\[ \Rightarrow \dfrac{{{m_y}}}{{{m_{He}}}} = \dfrac{{{v_{He}}}}{{{v_y}}}\]
\[ \Rightarrow \dfrac{{{v_{He}}}}{{{v_y}}} = \dfrac{{216}}{4} = 54\]
Using this, we can find the relation between kinetic energy of alpha particle and kinetic energy of decayed particle,
\[ \Rightarrow \dfrac{{K{E_y}}}{{K{E_{He}}}} = \dfrac{{\dfrac{1}{2}(216){v_y}^2}}{{\dfrac{1}{2}(4){v_{He}}^2}} = \dfrac{{54}}{{{{54}^2}}} = \dfrac{1}{{54}}\]
Thus, we found out the relation between the kinetic energy. Using this relation, substitute for any one value in equation (1)
\[ \Rightarrow K{E_y} + K{E_{He}} = 5MeV\]
We need to find KE of alpha particle, thus rearranging the equation we get
\[ \Rightarrow \dfrac{{K{E_{He}}}}{{54}} + K{E_{He}} = 5MeV\]
Taking LCM, we get
\[ \Rightarrow \dfrac{{55K{E_{He}}}}{{54}} = 5MeV\]
\[ \Rightarrow K{E_{He}} = \dfrac{{54}}{{11}}MeV\]

Thus , option (c) is the right answer for a given question.

Note Similar to alpha decay, there is a concept of beta decay. Beta decay is defined as a type of radioactive decay of an element where a positron (beta particle) is emitted by transforming the original particle into an isobar.