Radioactive decays - given power value

AI Thread Summary
The discussion revolves around calculating the energy available from the decay of uranium-238, its decay rate, and the relationship between power output and disintegration rates. Participants clarify that the decay rate can be derived from the half-life without additional quantities and emphasize the importance of unit consistency in calculations. Power is linked to the number of decays per second, which is crucial for determining activity in Curies. The conversation highlights common misconceptions about the relationship between power and the number of atoms involved in radioactive decay. Overall, the thread serves as a resource for understanding the calculations involved in radioactive decay and the significance of accurate unit conversions.
mahoteacher
Messages
3
Reaction score
0

Homework Statement


The masses of the following atoms are:

238 U → 238.050786(2)
234 Th→ 234.043598(5)
4 He → 4.002603

a) Calculate the energy available in the decay of 238U.

b) Calculate the decay rate of 238U

238U half time = 4.468*10^9yrs

c) If a block of uranium produces 1mW of power, how many disintegrations per second is this and what is the activity in Curies. Neglect other decays other than the one above.

d) Calculate the number of uranium atoms and the mass of the block.

- This is a question for an upper level medical physics class, but it seems to be relatively easy. So I am not sure if I am putting this in the right place.

Homework Equations


Mass calculation of radioactive decay

1 amu = 1.66 * 10^ -27 [kg]

t½ = ln2 / λ

A = λ*N = ΔN/ Δt

1 Ci = 3.7* 10^10 [s^-1]

The Attempt at a Solution


[/B]
a)
Δ amu = 238.050786(2) - 234.043598(5) - 4.002603 [amu] = 4.58*10^-3 [amu]

Δ mass = Δ amu * 1. 66*10^-27 [kg]
Δ mass = 4.58*10^-3 * 1.66 *10^-27 = 7.6106 *10^-30 [kg]

E = Δ mass * c^2
E = 7.6106 *10^-30 * 3*10^8 = 2.28 * 10^-21 J

b)

Δ t = 4.468*10^9 * 365 * 24 *60 * 60 = 1.409028* 10^17
Δ N = 6.022*10^23 atoms/ Δ amu = 7.9126* 10^-8

decay rate = ΔN/ Δt = 5.615 * 10^ -25

c)

disintegration per second = A = λN and 1 Ci = 3.7 * 10^10 [s^-1]

But how does power play a role? Does power affect the number of atoms? If yes how?
Assuming power has nothing to do with N, I am approaching the problem like this:

t½ = ln2/ λ
λ = 4.91932 * 10^ -18 [s^-1]
N = 6.022 * 10^ 23/ 238

A = λN = 12447 Bq = 3.36 * 10^ -7 [Ci]

d) since I don't know how power is related, I don't really know :(
 
Physics news on Phys.org
mahoteacher said:
E = Δ mass * c^2
E = 7.6106 *10^-30 * 3*10^8 = 2.28 * 10^-21 J
Where did the square of the speed of light go?
mahoteacher said:
Δ N = 6.022*10^23 atoms/ Δ amu = 7.9126* 10^-8
What did you calculate here? Working with units would make it easier to follow.
mahoteacher said:
decay rate = ΔN/ Δt = 5.615 * 10^ -25
That is not correct. You can directly get the decay rate from the half-life without other quantities involved.
mahoteacher said:
But how does power play a role? Does power affect the number of atoms? If yes how?
If every apple has a mass of 0.2 kilogram and I give you 5 kg per second ("power"), how many apples per second do you get?
Power does not "affect" the number of atoms but you need power to figure out how many decays happen per second.
 
Thank you for correcting me my mistakes, and helping me to move the thread to the right place!

E = Δ mass * c^2
E = 7.6106 *10^-30 *( 3*10^8)^2 = 6.84*10^-13 J

Δ N = 6.022*10^23 [atoms/mol] / Δ amu [g/mol] = 7.9126* 10^-8 [atoms/g]

Regarding the decay rate, I thought it means activity, which is A = λN. How can I get directly from half life? Can you give me some pointers?
t½ = ln2/ λ ==> λ = 4.91932 * 10^ -18 [s^-1]

For power, does that mean for question c)
Disintegration per second = A = 1*10^-3 [s^-1] ?

Thank you so much!
 
mahoteacher said:
Δ N = 6.022*10^23 [atoms/mol] / Δ amu [g/mol] = 7.9126* 10^-8 [atoms/g]
That would give 12 tons per atom (the inverse value). Very heavy atoms...

Decay rate: Hmm, could be a matter of definition. If you used it as decays per second per gram, then you need the number of atoms per gram, sure.
mahoteacher said:
For power, does that mean for question c)
Disintegration per second = A = 1*10^-3 [s^-1] ?
No. Where does that value come from?
 
mfb said:
That would give 12 tons per atom (the inverse value). Very heavy atoms...
Oh I see that doesn't make sense now...

mfb said:
Decay rate: Hmm, could be a matter of definition. If you used it as decays per second per gram, then you need the number of atoms per gram, sure.
So am I correct to think that for decay rate A = λN
t½ = ln2/ λ => λ = 4.91932 * 10^ -18 [s^-1]
Δ N = 6.022*10^23 [atoms/mol] / 238 [g / mol] = 2.530*10^21 [atoms/g]

A = 12447 [s^-1] = 1.24*10^4 [s^-1]

mfb said:
No. Where does that value come from?

That value comes from the power value 1 mW = 1*10^-3 W

This question seems really easy, but I haven't done physics in a while and couldn't really wrap my head around.
Thanks so much for your help!
 
mahoteacher said:
A = 12447 [s^-1] = 1.24*10^4 [s^-1]
Another value that comes out of the blue...
mahoteacher said:
That value comes from the power value 1 mW = 1*10^-3 W
W is not the same as 1/s.

All the quantities you need can be derived by combining given values in a way that the units match. If the units do not match, the answer is certainly wrong.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Replies
7
Views
3K
Replies
8
Views
7K
Replies
4
Views
2K
Replies
6
Views
5K
Replies
16
Views
3K
Back
Top