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.

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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 ${\mathrm{U}}^{238}\to {\mathrm{Th}}^{234}+{\mathrm{He}}^{4}$ 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 $E=∆m{c}^{2}\phantom{\rule{0ex}{0ex}}E=\left[0.00457\mathrm{u}\right]×931.5\mathrm{MeV}\phantom{\rule{0ex}{0ex}}⇒E=4.25467\mathrm{MeV}=4.255\mathrm{MeV}\phantom{\rule{0ex}{0ex}}$ (b) When two protons and two neutrons are emitted one by one, the reaction will be ${\mathrm{U}}^{233}\to {\mathrm{Th}}^{234}+2n+2p$ $\mathrm{Mass}\mathrm{defect},∆m=m\left({\mathrm{U}}^{238}\right)-\left[m\left({\mathrm{Th}}^{234}\right)+2\left({m}_{\mathrm{n}}\right)+2\left({m}_{p}\right)\right]\phantom{\rule{0ex}{0ex}}∆m=238.0508\mathrm{u}-\left[234.04363\mathrm{u}+2\left(1.008665\right)\mathrm{u}+2\left(1.007276\right)\mathrm{u}\right]\phantom{\rule{0ex}{0ex}}∆m=0.024712\mathrm{u}$ Energy released (E) when 238U emits two protons and two neutrons is given by $E=∆m{c}^{2}\phantom{\rule{0ex}{0ex}}E=0.024712×931.5\mathrm{MeV}\phantom{\rule{0ex}{0ex}}E=23.019=23.02\mathrm{MeV}$

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