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Unreasonable Results

  1. Repeat exercise but include the \({\rm{0}}{\rm{.0055 \% }}\) natural abundance of \(^{{\rm{234}}}{\rm{U}}\) with its \(2.45 \times {10^5}\) y half-life.
  2. What is unreasonable about this result?
  3. What assumption is responsible?
  4. Where does the \(^{{\rm{234}}}{\rm{U}}\) come from if it is not primordial?

Short Answer

Expert verified
  1. The value of \(\frac{{{N_0}}}{N} = {10^{5527}}\).
  2. Because the ratio is so high, the result is illogical.
  3. The implausible finding is due to the assumption that the \(^{{\rm{234}}}{\rm{U}}\)was not generated as a daughter nucleus and instead existed primordially.
  4. \(^{{\rm{234}}}{\rm{U}}\)is part of the \(^{{\rm{238}}}{\rm{U}}\)chain decay.

Step by step solution

01

Concept Introduction

The following is the relationship between activity, half-life, and the number of atoms:

\({\rm{R = }}\frac{{{\rm{0}}{\rm{.693N}}}}{{{{\rm{t}}_{{\rm{1/2}}}}}}\)

Where,

\({{\rm{t}}_{{\rm{1/2}}}}{\rm{ = }}\)Half life

\({\rm{R = }}\)Activity

\({\rm{N = }}\)Number of atoms

02

Value of \(\frac{{{{\rm{N}}_{\rm{0}}}}}{{\rm{N}}}\)

  1. The following is the relationship between the starting number and the final number of radioactive substances after time t:

\(N = {N_0}{e^{ - \lambda t}} \Rightarrow \frac{{{N_0}}}{N} = {e^{ - \lambda t}}\)

Where\({N_0} = \) initial activity,

\(N = \)final activity

\(\lambda = \frac{{0.693}}{{{t_{1/2}}}}\)and

\(t = \)time

\(\lambda = \frac{{0.693}}{{{t_{1/2}}}},{t_{1/2}} = 2.45 \times {10^5}y\)

Also,\({\rm{t = 4}}{\rm{.5 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{y }}\)

Substitute the values in the above equation,

\(\begin{aligned}\frac{{{N_0}}}{N} = {e^{\frac{{0.693\left( {4.5 \times {{10}^9}y} \right)}}{{2.45 \times {{10}^5}y}}}}\\ = {e^{12728}}\\ = {10^{5527}}\end{aligned}\)

Therefore, the value of \(\frac{{{N_0}}}{N} = {10^{5527}}\).

03

Unreasonable about this result

  1. Because the ratio is so high, the result is illogical.
04

Explanation of  assumption

  1. The implausible finding is due to the assumption that the \(^{{\rm{234}}}{\rm{U}}\) was not generated as a daughter nucleus and instead existed primordially.
05

\(^{{\rm{234}}}{\rm{U}}\) is a part of chain decay

  1. The \(^{{\rm{234}}}{\rm{U}}\) is part of the \(^{{\rm{238}}}{\rm{U}}\) chain decay.

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Most popular questions from this chapter

A \({}^{{\rm{60}}}{\rm{Co}}\) source is labelled \(4.00\,{\rm{mCi}}\), but its present activity is found to be \(1.85 \times {10^7}\,{\rm{Bq}}\). (a) What is the present activity in \({\rm{mCi}}\)? (b) How long ago did it actually have a \({\rm{4}}{\rm{.00mCi}}\) activity?

Find the length of a side of a cube having a mass of \(1.0\,{\rm{kg}}\) and the density of nuclear matter, taking this to be \(2.3 \times {10^{17}}\,{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\).

Unreasonable Results

The manufacturer of a smoke alarm decides that the smallest current of \({\rm{\alpha }}\) radiation he can detect is \(1.00\,\mu A\).

  1. Find the activity in curies of an \({\rm{\alpha }}\) emitter that produces a \(1.00\,\mu A\)current of \({\rm{\alpha }}\) particles.
  2. What is unreasonable about this result?
  3. What assumption is responsible?

Consider the generation of electricity by a radioactive isotope in a space probe, such as described in exercise. Construct a problem in which you calculate the mass of a radioactive isotope you need in order to supply power for a long space flight. Among the things to consider are the isotope chosen, its half-life and decay energy, the power needs of the probe and the length of the flight.

What is the ratio of the velocity of a \(5.00\,{\rm{MeV }}\beta \)ray to that of an\({\rm{\alpha }}\) particle with the same kinetic energy? This should confirm that \({\rm{\beta }}\) s travel much faster than \({\rm{\alpha }}\) s even when relativity is taken into consideration. (See also Exercise \({\rm{31}}{\rm{.11}}\).)

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