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

[Rectifier]

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?

 

[Metals]

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.

 

[mfb]

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.

 

[Rectifier]

Approximately, but looking that up is certainly not the intended solution if the half-life is given.

@Rectifier: Correct.

Thank you!

 

[Bystander]

Is that solution correct?

Works.

 

[Metals]

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.

 

[insightful]

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?

 

[mfb]

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.

 

[Rectifier]

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?

 

[insightful]

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!

 

[Rectifier]

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

 

[insightful]

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.

 

[mfb]

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? :)