Question

Only 235u can be used as fuel in a nuclear reactor, so uranium for use in the nuclear industry must be enriched in this isotope. if a sample of enriched uranium has an average atomic mass of 236.253 amu, what percentage of 235u is present?

Answer 1

Enriched uranium is used as fuel in nuclear reactors by increasing the percentage of uranium-235. In the case of a sample with an average atomic mass of 236.253 amu, approximately 99.9% of the uranium in the sample is uranium-235.

Step-by-step explanation:

Enriched uranium is used as fuel in nuclear reactors by increasing the percentage of uranium-235. In its natural form, uranium consists of more than 99% uranium-238 and less than 1% uranium-235. To make it suitable for use in a nuclear reactor, the concentration of uranium-235 is increased to around 3-5%. This enrichment process separates the uranium isotopes based on their atomic masses using techniques like centrifugation or gaseous diffusion.

In the case of a sample of enriched uranium with an average atomic mass of 236.253 amu, we need to find the percentage of uranium-235 present in it. Since uranium-238 has an atomic mass of 238.050 amu, and uranium-235 has an atomic mass of 235.044 amu, we can determine the amount of uranium-235 by subtracting the atomic mass of uranium-238 from the average atomic mass of the sample.

By subtracting 238.050 amu from 236.253 amu, we get 235.203 amu. Dividing this by the atomic mass of uranium-235 (235.044 amu) gives us a ratio of 0.999; this indicates that 99.9% of the uranium in the sample is uranium-235. Therefore, the percentage of uranium-235 in the enriched uranium sample is approximately 99.9%.

Answer 2

Approximately 59.7739%.

Uranium has two naturally-occurring isotopes (note that despite the fact that both species are radioactive, they are the most abundant ones for having a relatively-long half-life.)

• Uranium 235 with a relative atomic mass of
  `235.0439299 \; \text{amu}` and
• Uranium 238 with a relative atomic mass of
  `238.05078826\; \text{amu}`

Assuming that
`x \; \%` of the uranium atoms in this sample are atoms of
`^(235)\text{U}`; given the fact that other isotopes of uranium contribute to a negligible share of mass of the sample (less than
`0.01\%` in naturally-occurring samples,)
`^(238)\text{U}` would have contributed to
`1-x\; \%` of all atoms in this sample.

Similar to the standard atomic mass for an element, the average atomic mass of this sample shall resembles the atomic mass weighted over all atoms in the collection. Therefore

`\%^(235)\text{U Abundancy} * m_a(^(235)\text{U}) + \%^(238)\text{U Abundancy} * m_a(^(238)\text{U}) = a_r(\text{sample})`

`235.043 \; x + 238.051 \; (1-x) = 236.253\\ x \approx 0.597739 = 59.7739 \; \%`