# Time taken for amount of a nuclear material to remain

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1. Dec 17, 2015

### InsaneScientist

1. The problem statement, all variables and given/known data
The half life T½ of Carbon-14 is 5730 years. What is its decay constant? After what length of time will 35% of an initial sample of Carbon-14 remain?

2. Relevant equations
Decay constant λ= 0.693 / T½
Where N = amount of radioactive substance,
N=N0e-λt

3. The attempt at a solution
Okay, so I can get the decay constant easily,
λ= 0.693 / 5730 = 1.2094x10-4
but I just can't figure out an expression for the time taken for the amount of nuclear material to remain.
I'm thinking the amount can be expressed in the formula:
0.35=e-λt
It's probably some simple algebra but I can't figure out how to isolate t in this equation

2. Dec 17, 2015

### Mister T

Have you reviewed the basic algebra associated with the exponential function? I'll give you a hint: logarithm.

3. Dec 17, 2015

### SteamKing

Staff Emeritus
Remember, ln (ex) = x

4. Dec 17, 2015

### InsaneScientist

Thanks for the hint :) I haven't done any maths in over 2 years before starting my physics course so I've forgotten a lot of basic stuff which they tend to skip out on in Uni.

logeN = -λt
loge0.35 = -1.209x10-4t

t = loge0.35 / -1.209x10-4 = 8683.4 years.
Is this correct?

5. Dec 17, 2015

### SteamKing

Staff Emeritus
Seems to be. Half of the C-14 disappears after 5730 years, and 75% is gone after 5730 + 5730 = 11,460 years.

6. Dec 18, 2015

### SammyS

Staff Emeritus
You can also find the number of half-lifes to get 35% by solving $\displaystyle \ 0.35=\left(\frac{1}{2}\right)^x \,,\$ where x is the number of half-lifes .

That gives an answer close to yours.
8678.5 years