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Given specimens uranium\({\rm{ - 232}}\)(\({{\rm{t}}_{{\rm{1/2}}}}{\rm{ = 68}}{\rm{.9 y}}\)) and uranium\({\rm{ - 233}}\)(\({{\rm{t}}_{{\rm{1/2}}}}{\rm{ = 159,200 y}}\)) of equal mass, which one would have greater activity and why?

Short Answer

Expert verified

The activity is inversely proportional to the half-life, which indicates that the higher the half-life, the lower the activity. Uranium\({\rm{ - 232}}\) has a shorter half-life than uranium\({\rm{ - 233}}\), which means it has higher activity.

Step by step solution

01

Step 1: Nuclear Chemistry

Nuclear chemistry is a branch of chemistry that studies radioactivity, nuclear processes, and atomic nuclei alterations such as nuclear transmutation and nuclear characteristics.

02

Explanation

The next formula is used here for this task as:

\({\rm{t = }}\frac{{\rm{1}}}{{\rm{\lambda }}}{\rm{In}}\frac{{{{\rm{n}}_{\rm{0}}}}}{{{{\rm{n}}_{\rm{t}}}}}\)

As the atomic masses of both substances are the same, the nuclide concentration will be the same.

The decay constant has a formula as well:

\({\rm{\lambda = }}\frac{{{\rm{In(2)}}}}{{{{\rm{t}}_{{\rm{1/2}}}}}}\)

As there are two isotopes, there is an expression for determining their activity:

\({\rm{activity = \lambda \times N}}\)

We may put the decay constant from above in that formula:

\({\rm{activity = }}\frac{{{\rm{In(2)}}}}{{{{\rm{t}}_{{\rm{1/2}}}}}}{\rm{ \times N}}\)

As a result, we may deduce that activity and half-life are inversely proportional:

\({\rm{activity \alpha }}\frac{{{\rm{In(2)}}}}{{{{\rm{t}}_{{\rm{1/2}}}}}}\)

If the half-life is longer, the activity will be lower. Uranium\({\rm{ - 232}}\) has a shorter half-life than uranium\({\rm{ - 233}}\), which means it has higher activity.

Therefore, the activity is said to be inversely proportional to half-life. It means that the activity will be smaller with a higher half-life. Uranium\({\rm{ - 232}}\) has to be a smaller half-life, and thus it has more activity than uranium\({\rm{ - 233}}\).

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Most popular questions from this chapter

Question:15. Write a balanced equation for each of the following nuclear reactions:

  1. The product of \(^{{\bf{17}}}{\bf{O}}\) from \(^{{\bf{14}}}{\bf{N}}\) by \({\bf{\alpha }}\) particle bombardment
  2. The production of \(^{{\bf{14}}}{\bf{C}}\) from \(^{{\bf{14}}}{\bf{N}}\) by neutron bombardment
  3. The production of \(^{{\bf{233}}}{\bf{Th}}\) from \(^{{\bf{232}}}{\bf{Th}}\) by neutron bombardment
  4. The production of \(^{{\bf{239}}}{\bf{U}}\) from \(^{{\bf{238}}}{\bf{U}}\) by \({_{\bf{1}}^{\bf{2}}}{\bf{H}}\) bombardment\(\)

Question: Explain, in terms of Figure \({\rm{21}}{\rm{.2}}\), how unstable heavy nuclides (atomic number \({\rm{ > 83}}\)) may decompose to form nuclides of greater stability

(a) if they are below the band of stability and

(b) if they are above the band of stability.

Question: The following nuclei do not lie in the band of stability. How would they be expected to decay? Explain your answer.

(a) \({}_{{\rm{15}}}^{{\rm{34}}}{\rm{P}}\)

(b) \({}_{{\rm{92}}}^{{\rm{239}}}{\rm{U}}\)

(c) \({}_{20}^{{\rm{38}}}{\rm{Ca}}\)

(d) \({}_{\rm{1}}^{\rm{3}}{\rm{H}}\)

(e) \({}_{94}^{{\rm{245}}}{\rm{Pu}}\)

Write the balanced nuclear equation for the production of the following transuranium elements:

(a) berkelium-244, made by the reaction of Am-241 and He-4

(b) fermium-254, made by the reaction of Pu-239 with a large number of neutrons

(c) lawrencium-257, made by the reaction of Cf-250 and B-11

(d) dubnium-260, made by the reaction of Cf-249 and N-15

Question: What is the change in the nucleus that results from the following decay scenarios?

(a) emission of a \({\rm{\beta }}\) particle

(b) emission of a \({{\rm{\beta }}^{\rm{ + }}}\) particle

(c) capture of an electron

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