Half life and percentage remaining

[nmsurobert]

Homework Statement

¹³⁷Cs has a half life t_1/2 = 30.07 years. After 50 years, what fraction of ¹³⁷Cs will remain?

Homework Equations

maybe...
dN/dλ = λN
λ = ln2/t_1/2, half life in seconds
N = Total mass /(137(1.67x10^-17kg))

The Attempt at a Solution

i thought maybe i needed to figure out a decay rate at first but i don't have a total mass. unless i just use the mass of the cesium. so my N will simply equal 1/(1.67x10^-17)

but that gives me a huge number. 43 million decays per second.

so then i thought if i lose 50% in 30 years. then in 20 years, which is 66% the amount of time needed for another half life to pass, then i should lose 66% of the 50% which is only 33%. so in 50 years i lose 33% of the 50% lost in the first 30 years. but then i get lost in my logic and don't know how to proceed.

edit: if i take .50 - .33 = .17. then add the .17 to the .50 i get .67. this makes sense to me as after two half lives have passed then .75 of the Cs will be left over.

 

[Chestermiller]

Homework Statement

¹³⁷Cs has a half life t_1/2 = 30.07 years. After 50 years, what fraction of ¹³⁷Cs will remain?

Homework Equations

maybe...
dN/dλ = λN

This equation is incorrect. It should read:

$$\frac{dN}{dt}=-\lambda N$$

where
λ = ln2/t_1/2, half life in seconds

If t_1/2 = 30.07 years, from this equation, what is the value of λ?

If N = N₀ at time t = 0, do you know how to solve the above differential equation for the value of N at arbitrary time t (in terms of N₀, λ, and t)?

Chet

 

[nmsurobert]

i do not know how to solve that equation.

but what i did was 50/30.07 = 1.66. and if 2^-1 is .5. then 2^-1.66 is .68. which leave 32% of the cesium remaining.

 

[haruspex]

i do not know how to solve that equation.

but what i did was 50/30.07 = 1.66. and if 2^-1 is .5. then 2^-1.66 is .68. which leave 32% of the cesium remaining.

If you do not know how to solve that equation, and are not expected to be able to solve it, then you must have been given another relevant equation. Maybe one involving e?

 

[Chestermiller]

i do not know how to solve that equation.

but what i did was 50/30.07 = 1.66. and if 2^-1 is .5. then 2^-1.66 is .68. which leave 32% of the cesium remaining.

Yes. I agree with haruspex. On what mathematical basis did you choose this expression over the infinite number of other possibilities available?

Chet