Is Radioactive Decay Calculation Confusing?

  • Thread starter Sanosuke Sagara
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In summary, the conversation discusses a question and solution provided in an attachment. The solution includes equations and calculations for various parts of the question, including finding the amount of radioactive material after a certain time and determining the time needed for a second treatment. The conversation also includes comments and feedback on the solution.
  • #1
Sanosuke Sagara
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I have my doubt , solution and the question in the attachment that followed.
Thanks for anybody that spend some time on this question.
 

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  • Doc 5 (Radioactive).doc
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  • #2
Sanosuke Sagara said:
I have my doubt , solution and the question in the attachment that followed.
Thanks for anybody that spend some time on this question.

I think you made some careless errors.

Your answer for a is right.

I'd like to write out the equations...
[tex]\frac{-dn}{dt} = 5000(\frac{1}{2})^{t/8}[/tex] where t is in hours

Just plug in t=1 into the right side and you get your answer for a. 4585

If we integrate both sides we get

[tex]n(t) = -\frac{5000}{ln(1/2)}(\frac{1}{2})^{t/8}[/tex] or

[tex]n(t) = \frac{5000}{ln(2)} (\frac{1}{2})^{t/8} [/tex]

Now since the time in the equations are hours... there's no need to convert to seconds.

n(0) = 5000/ln(2) = 7213
n(1) = (5000/ln(2)) (1/2)^(1/8) = 6615

So for b) n(0) - n(1) = 598

c) She comes back 24 hours later. So in the equation we'll need to plug in t=25 (24 hours after t=1) to see how much of the radioactive stuff is left.
n(25) = (5000/ln(2)) (1/2)^(25/8) = 827

To get the same treatment she needs to get 598 atoms just like part b). So the total needs to come down from 827 to 827-598=229 after the second treatment is finished

We solve n(t) = 229, I get t=40 hours.

So the second treatment starts at t=25 hours, and finished t=40 hours to get 15 hours of treatment.

Check the work yourself. I might have made some careless errors.
 
Last edited:
  • #3
Thanks for your explanation and comment to me for my unneccasary confusing way of calculation.I think I have understand what the question want and I will try to find out the answer.Thanks again,learningphysics.
 

Related to Is Radioactive Decay Calculation Confusing?

1. What is radioactivity?

Radioactivity is the spontaneous emission of particles or energy from an unstable nucleus of an atom. This process is also known as radioactive decay or nuclear decay.

2. What are the types of radiation emitted during radioactive decay?

The three main types of radiation emitted during radioactive decay are alpha particles, beta particles, and gamma rays. Alpha particles are two protons and two neutrons bound together, beta particles are high-energy electrons or positrons, and gamma rays are high-frequency electromagnetic waves.

3. How is radioactivity measured?

The unit of measurement for radioactivity is the becquerel (Bq), which represents one radioactive decay per second. Another commonly used unit is the curie (Ci), where 1 Ci equals 3.7 x 10^10 Bq.

4. What are the risks associated with exposure to radioactivity?

The risks associated with exposure to radioactivity depend on the type, energy, and duration of exposure. High levels of exposure can cause acute health effects such as radiation sickness, while chronic exposure to low levels can increase the risk of developing cancer.

5. How is radioactivity used in everyday life?

Radioactivity has many practical uses in everyday life, such as in medical imaging and treatment, industrial applications, and energy production. It is also used in smoke detectors, food preservation, and carbon dating in archaeology.

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