Question

Consider the following nuclear fission reaction

${}_{88}R{a}^{226}\to {}_{86}R{n}^{222}+{}_{2}H{e}^4+Q$

In this fission reaction. the kinetic energy of α-particle emitted is $4.44Mev$. Find the energy emitted as γ – radiation in $KeV$ in this reaction.

Answer 1

Step 1. Given data

${}_{88}R{a}^{226}\to {}_{86}R{n}^{222}+{}_{2}H{e}^4+Q$

Kinetic energy of $\alpha$ particle, ${\left(K.E.\right)}_{\alpha }=4.44Mev$

Step 2. Finding the $Q$ value

Mass of ${}_{88}R{a}^{226}$, $m_1=226.005amu$

Mass of ${}_{86}R{n}^{222},m_2=222.000amu$

Mass of ${}_{2}H{e}^4,m_3=4amu$

Mass defect, $∆m$

,$$\begin{gathered} ∆m=m_1-(m_2+m_3) \\ ∆m=226.005-\left(222.000+4\right) \\ ∆m=0.005amu \end{gathered}$$

By using the formula of the $Q$-value

$$\begin{gathered} Q-value=∆m\times 931.5Mev \\ =0.005\times 931.5 \\ =4.655Mev \end{gathered}$$

Step 3. Finding the energy emitted, $E$

As kinetic energy is directly proportional to the mass.

By using the formula,

$$\begin{gathered} \frac{{\left(K.E.\right)}_{Rn}}{{\left(K.E.\right)}_{\alpha }}=\frac{m_3}{m_2} \\ {\left(K.E.\right)}_{Rn}=\frac{m_3}{m_2}\times {\left(K.E.\right)}_{\alpha } \\ =\frac{4}{222}\times 4.44 \\ {\left(K.E.\right)}_{Rn}=0.08Mev \end{gathered}$$ [${\left(K.E.\right)}_{Rn}$ is the kinetic energy of $Rn$]

By using the formula of energy,

$$\begin{gathered} E=Q-\left({\left(K.E.\right)}_{Rn}+{\left(K.E.\right)}_{\alpha }\right) \\ E=4.655-\left(0.08+4.44\right) \\ E=0.135Mev \\ E=0.135\times {10}^{3}Kev \end{gathered}$$

Therefore, the energy emitted is $0.135\times {10}^{3}Kev$.