Uranium- 235 undergoes many different fission reactions. For one such reaction, when \({ }^{235} \mathrm{U}\) is struck with a neutron, \({ }^{144} \mathrm{Ce}\) and \({ }^{90} \mathrm{Sr}\) are produced along with some neutrons and electrons. How many neutrons and \(\beta\) -particles are produced in this fission reaction?

Short Answer

Expert verified
In the fission reaction of Uranium-235, when struck with a neutron to form Cerium-144 and Strontium-90, 2 neutrons and 4 beta particles are produced.

Step by step solution

01

Write down the given fission reaction

Let's start by writing down the nuclear fission reaction of Uranium-235 being struck with a neutron: \[_{0}^1n + _{92}^{235}U \rightarrow _{58}^{144}Ce + _{38}^{90}Sr + x_{0}^1n + y_{-1}^0\beta\] Here, \(x\) represents the number of neutrons and \(y\) represents the number of beta particles produced in the reaction.
02

Apply conservation of mass number

In this step, we will apply the conservation of mass number to the given reaction. The sum of the mass numbers on the left side of the reaction must equal the sum of the mass numbers on the right side. Mass number on the left side = Mass number on the right side \(1 + 235 = 144 + 90 + x\) Now, let's solve for \(x\): \(x = 1 + 235 - 144 - 90\) \(x = 2\) This tells us that 2 neutrons are produced in the fission reaction.
03

Apply conservation of atomic number

Next, we will apply the conservation of atomic number to the given reaction. The sum of the atomic numbers on the left side of the reaction must equal the sum of the atomic numbers on the right side. Atomic number on the left side = Atomic number on the right side \(0 + 92 = 58 + 38 - y\) Now, let's solve for \(y\): \(-y = 0 + 92 - 58 - 38\) \(y = -4\) Since a negative value for the number of beta particles does not make sense, we made an error when writing the fission reaction. In beta decay, beta particles are emitted and atomic number increases by 1. Therefore, the equation should be written as: \[_{0}^1n + _{92}^{235}U \rightarrow _{58}^{144}Ce + _{38}^{90}Sr + x_{0}^1n + y_{+1}^0\beta\] Applying conservation of atomic number again: \(0 + 92 = 58 + 38 + y\) Solve for \(y\): \(y = 0 + 92 - 58 - 38\) \(y = 4\) This tells us that 4 beta particles are produced in the fission reaction.
04

Conclusion

In summary, the given fission reaction of Uranium-235 produces 2 neutrons and 4 beta particles.

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