Question

(a) Calculate the energy released if ²³⁸U emits an α-particle. (b) Calculate the energy to be supplied to ²³⁸U it two protons and two neutrons are to be emitted one by one. The atomic masses of ²³⁸U, ²³⁴Th and ⁴He are 238.0508 u, 234.04363 u and 4.00260 u respectively.

Answer 1

(a)
Given:
Atomic mass of ²³⁸U, m(²³⁸U) = 238.0508 u
Atomic mass of ²³⁴Th, m(²³⁴Th) = 234.04363 u
Atomic mass of ⁴He, m(⁴He) = 4.00260 u
When ²³⁸U emits an α-particle, the reaction is given by
${\mathrm{U}}^{238}\to {\mathrm{Th}}^{234}+{\mathrm{He}}^4$
Mass defect, Δm = [m(²³⁸U)$-$(m(²³⁴Th)+m(⁴He))]
Δm = [238.0508 $-$ (234.04363 + 4.00260) = 0.00457 u
Energy released (E) when ²³⁸U emits an α-particle is given by
$$\begin{gathered} E=∆m{c}^2 \\ E=[0.00457\mathrm{u}]\times 931.5\mathrm{MeV} \\ \Rightarrow E=4.25467\mathrm{MeV}=4.255\mathrm{MeV} \end{gathered}$$

(b)

When two protons and two neutrons are emitted one by one, the reaction will be
${\mathrm{U}}^{233}\to {\mathrm{Th}}^{234}+2n+2p$

$$\begin{gathered} \mathrm{Mass}\mathrm{defect},∆m=m\left({\mathrm{U}}^{238}\right)-[m\left({\mathrm{Th}}^{234}\right)+2\left(m_{\mathrm{n}}\right)+2\left(m_p\right)] \\ ∆m=238.0508\mathrm{u}-[234.04363\mathrm{u}+2(1.008665)\mathrm{u}+2(1.007276)\mathrm{u}] \\ ∆m=0.024712\mathrm{u} \end{gathered}$$

Energy released (E) when ²³⁸U emits two protons and two neutrons is given by

$$\begin{gathered} E=∆m{c}^2 \\ E=0.024712\times 931.5\mathrm{MeV} \\ E=23.019=23.02\mathrm{MeV} \end{gathered}$$