#### Question

(a) Calculate the energy released if ^{238}U emits an α-particle. (b) Calculate the energy to be supplied to ^{238}U it two protons and two neutrons are to be emitted one by one. The atomic masses of ^{238}U, ^{234}Th and ^{4}He 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/c^{2},1 u = 931 MeV/c^{2}.)

#### Solution

(a)

Given:

Atomic mass of ^{238}U, m(^{238}U) = 238.0508 u

Atomic mass of ^{234}Th, m(^{234}Th) = 234.04363 u

Atomic mass of ^{4}He, m(^{4}He) = 4.00260 u

When ^{238}U 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"`