Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# (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. - Physics

Sum

(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 mp = 1.007276 u, Mass of ""_1^1"H" atom = 1.007825 u, Mass of neutron mn = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c2,1 u = 931 MeV/c2.)

#### Solution

(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"

Concept: Mass-energy and Nuclear Binding Energy - Mass - Energy
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 24 The Nucleus
Q 6 | Page 442