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.

(Use Mass of proton m_p = 1.007276 u, Mass of `""_1^1"H"` atom = 1.007825 u, Mass of neutron m_n = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c²,1 u = 931 MeV/c².)

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

`"U"^238 → "Th"^234 + "He"^4`

Mass defect , `Δm = [m(""^238U - (m(""^234"Th") + m(""^4He))]`

`Δm = [238.0508 - (234.04363 + 4.00260)] = 0.00457 "u"`

Energy released (E) when `""^238U` emits an α-particle is given by 

`E = Δm  c^2`

`E = [0.00457  "u"] xx 931.5  "MeV"`

⇒ `E = 4.25467  "MeV" = 4.255  "MeV"`

 

(b)

When two protons and two neutrons are emitted one by one, the reaction will be 

`"U"^233 → "Th"^234 + 2n + 2p`

Mass defect , `Δm = m("U"^238) - [m("Th"^234) + 2("m"_n) + 2(m_p)]`

`Δm = 238.0508  "u" - [234.04363  "u" + 2(1.008665) "u" + 2(1.007276) "u"]`

`Δm = 0.024712  "u"`

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

`E = Δmc^2`

`E = 0.024712 xx 931.5  "MeV"`

`E = 23.019 = 23.02  "MeV"`