# Yes, your solutions for both parts (a) and (b) are correct. Good job!

• Rectifier
In summary,The problem statement is translated from Swedish.There is an activity of 1μg Cobalt-60 in Bq (decays/second).You find a sample of Cobalt-60 that has a mass of 10 µg and has an activity of 5.0MBq. How old is that sample?The attempt at a solution is to calculate the activity of 1 microgram of Cobalt-60 using the N A constant.The solution is correct if the half-life is given.
Rectifier
Gold Member
The problem statement
The Cobalt-60 isotope is often used in medicine. It has a half-life of 5.25 years.
a) What is the activity of 1μg Cobalt in Bq (decays/second)?
b) You find a sample of Cobalt that has a mass of 10 µg and has an activity of 5.0MBq. How old is that sample?

This problem was translated from Swedish.

The attempt at a solution
a)
Activity is given by:
## A_{Bq}= n N_A \frac{ln(2)}{t_{\frac{1}{2}}} ##
where ## t_{\frac{1}{2}} = 1.656 \cdot 10^8 ## since ## 5.25 years =1.656 \cdot 10^8 seconds ## and ##N_A## is avogados constant which is ## 6.022×10^{23} mol^{−1} ## but what is n?

I guess that that is the amount of moles in 1μg Cobalt. Wolfram tells me its
## 1.6968366×10^{-8} mol ##

Thus the activity is
## A_{Bq}= n N_A \frac{ln(2)}{t_{\frac{1}{2}}} = 1.6968366 \cdot 10^{-8}\cdot 6.022\cdot10^{23}\cdot \frac{ln(2)}{1.656 \cdot 10^8}=
4.27707 \cdot 10^7##

b)
## A(t) = A_0 0.5^{\frac{t}{ t_{\frac{1}{2}} }} ##
where
A(t) = 5.0MBq
A(0) is the activity of 1μg Cobalt
##t_{\frac{1}{2}}## = ## 1.656 \cdot 10^8 ##

## A(t) = A_0 0.5^{\frac{t}{ t_{\frac{1}{2}} }} \\ 5.0 \cdot 10^6 = 4.27707 \cdot 10^7 \cdot 0.5^{\frac{t}{ 1.656 \cdot 10^8} } \\ t =5.12801×10^8 s ##
Which is about 16.26 years.

Is that solution correct?

Last edited:
The activity of 1 gram of Cobalt-60 is 44 TBq, so 44 trillion Bq. The activity of a μg (microgram) of Cobalt-60 would be 44,000,000,000,000 divided by 1,000, which is 44,000,000,000 (44 billion Bq). So the answer to part a): The activity of 1 microgram of Cobalt-60 is 44 billion (44,000,000,000) Bq.

I'm afraid I can't help with part b) though. Hope this helped.

Metals said:
The activity of 1 gram of Cobalt-60 is 44 TBq, so 44 trillion Bq.
Approximately, but looking that up is certainly not the intended solution if the half-life is given.
Also note that a gram has 1 million microgram, not 1000.

@Rectifier: Correct.

Metals and Rectifier
mfb said:
Approximately, but looking that up is certainly not the intended solution if the half-life is given.

@Rectifier: Correct.
Thank you!

Rectifier said:
Is that solution correct?
Works.

Rectifier
mfb said:
Approximately, but looking that up is certainly not the intended solution if the half-life is given.
Also note that a gram has 1 million microgram, not 1000.

@Rectifier: Correct.

Apologies, thanks for the correction.

For part a), did you find the number of moles for Co-60 or Co-59 (natural cobalt)?

For part b), should A0 be for 1 ug or 10 ug?

Rectifier and mfb
insightful said:
For part a), did you find the number of moles for Co-60 or Co-59 (natural cobalt)?

For part b), should A0 be for 1 ug or 10 ug?
Good points. I should have checked the numbers more carefully.

insightful said:
For part a), did you find the number of moles for Co-60 or Co-59 (natural cobalt)?

For part b), should A0 be for 1 ug or 10 ug?
a) that was the natural cobalt its not specified in the problem :(
b) ah darn, I guess I have to recalculate A_0 for 10 micro-grams. Is it just 10 times bigger by any chance?

Rectifier said:
a) that was the natural cobalt its not specified in the problem :(
b) ah darn, I guess I have to recalculate A_0 for 10 micro-grams. Is it just 10 times bigger by any chance?
Bingo!

I am not entierly sure about what to do with a) thoug? Any tips?

Rectifier said:
I guess that that is the amount of moles in 1μg Cobalt. Wolfram tells me its
1.6968366×10−8mol 1.6968366×10^{-8} mol
Just correct this using the result for M from your other thread.

Rectifier
Rectifier said:
a) that was the natural cobalt its not specified in the problem :(
It is, you have pure Co-60.

Rectifier
Attempt 2 :D

a)

## A_{Bq}= n N_A \frac{ln(2)}{t_{\frac{1}{2}}} ##
## t_{\frac{1}{2}} = 1.656 \cdot 10^8 ##
##N_A = 6.022 \cdot 10^{23} mol^{−1} ##
## n = \frac{n}{M} = \frac{10^{-6}}{59.933817059} = 1.668507 \cdot 10^{-8} ##

## A_{Bq}= n N_A \frac{ln(2)}{t_{\frac{1}{2}}} = 1.668507 \cdot 10^{-8} \cdot 6.022 \cdot 10^{23} \cdot \frac{ln(2)}{1.656 \cdot 10^8}= 4.20566 \cdot 10^7##b)
## A(t) = A_0 0.5^{\frac{t}{ t_{\frac{1}{2}} }} ##
where
A(t) = 5.0MBq
##t_{\frac{1}{2}} = 1.656 \cdot 10^8 ##

A(0) is the activity of 10μg Cobalt
## n = \frac{n}{M} = \frac{10 \cdot 10^{-6}}{59.933817059} = 1.6685 \cdot 10^{-7}## (turns out activity of 1μg * 10 = activity of 10μg)

## A_{Bq} = n N_A \frac{ln(2)}{t_{\frac{1}{2}}} \\ A_{Bq} = 1.6685 \cdot 10^{-7} 6.022 \cdot 10^{23} \frac{ln(2)}{1.656 \cdot 10^8} = \\ A_{Bq} = 4.20564 \cdot 10^8##

## A(t) = A_0 0.5^{\frac{t}{ t_{\frac{1}{2}} }} \\ 5.0 \cdot 10^6 = 4.20564 \cdot 10^8 \cdot 0.5^{\frac{t}{ 1.656 \cdot 10^8 }} = 1.05889 \cdot 10^9 s ##
Or also 33.58 years

Is this right then? :)

## 1. What is cobalt decay?

Cobalt decay is the process in which the unstable radioactive isotope cobalt-60 (Co-60) emits radiation and transforms into a more stable isotope over time.

## 2. How does cobalt decay occur?

Cobalt decay occurs through the emission of gamma radiation, which is a type of electromagnetic radiation. As the Co-60 isotope decays, it releases high-energy gamma rays, which can be dangerous if not properly contained.

## 3. What is the half-life of cobalt-60?

The half-life of cobalt-60 is approximately 5.27 years. This means that after 5.27 years, half of the initial amount of Co-60 will have decayed into a more stable isotope.

## 4. How is cobalt decay used in medicine?

Cobalt-60 is commonly used in medical treatments, particularly in radiation therapy for cancer. The high-energy gamma rays emitted during cobalt decay can be directed at cancerous cells to destroy them.

## 5. What safety precautions should be taken when working with cobalt decay?

Working with cobalt decay requires strict safety measures to protect against the harmful effects of radiation. These can include wearing protective gear, following proper handling procedures, and ensuring that all equipment and materials are properly shielded.

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