Write equations for the following nuclear reactions. (a) bombardment of \(^{238} \mathrm{U}\) with \(\alpha\) particles to produce \(^{239} \mathrm{Pu}\) (b) bombardment of tritium ( \(^{3} \mathrm{H}\) ) with \(_{1}^{2} \mathrm{H}\) to produce \(^{4} \mathrm{He}\) (c) bombardment of \(^{33} \mathrm{S}\) with neutrons to produce \(^{33} \mathrm{P}\).

Short Answer

Expert verified
The nuclear reactions are: (a) \(^{238}_{92} \mathrm{U} + ^{4}_{2} \mathrm{He} \rightarrow ^{239}_{94} \mathrm{Pu} + ^{1}_{0}n\),(b) \(^{3}_{1} \mathrm{H} + _{1}^{2} \mathrm{H} \rightarrow ^{4}_{2} \mathrm{He} + ^{1}_{0}n\),(c) \(^{33}_{16} \mathrm{S} + ^{1}_{0}n \rightarrow ^{33}_{15} \mathrm{P} + ^{-1}_{0}e\).

Step by step solution

01

Write Equation for Reaction a

For the reaction \(^{238} \mathrm{U} + \alpha \rightarrow ^{239} \mathrm{Pu} + ?\), An alpha particle has an atomic number of 2 and a mass of 4. To make the atomic numbers and masses balance, the particle produced should have an atomic number of 2 and a mass of 3. Thus, the product of the reaction is a neutron \(^{1}_{0}n\), so the reaction is \(^{238}_{92} \mathrm{U} + ^{4}_{2} \mathrm{He} \rightarrow ^{239}_{94} \mathrm{Pu} + ^{1}_{0}n\)
02

Write Equation for Reaction b

Consider the reaction \(^{3} \mathrm{H} + _{1}^{2} \mathrm{H} \rightarrow ^{4} \mathrm{He} + ?\). The missing particle must have an atomic number of 0 and a mass number of 1 to make the atomic numbers and masses balance, which is a neutron \(^{1}_{0}n\). So the reaction is \(^{3}_{1} \mathrm{H} + _{1}^{2} \mathrm{H} \rightarrow ^{4}_{2} \mathrm{He} + ^{1}_{0}n\)
03

Write Equation for Reaction c

For the reaction \(^{33} \mathrm{S} + neutron \rightarrow ^{33} \mathrm{P} + ?\), because the atomic numbers and masses must balance, the missing particle would essentially be an electron \(^{-1}_{0}e\) So the reaction is \(^{33}_{16} \mathrm{S} + ^{1}_{0}n \rightarrow ^{33}_{15} \mathrm{P} + ^{-1}_{0}e\)

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