Time taken for amount of a nuclear material to remain

[InsaneScientist]

Homework Statement

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?

Homework Equations

Decay constant λ= 0.693 / T½
Where N = amount of radioactive substance,
N=N₀e^-λt

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

 

[Herman Trivilino]

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

 

[SteamKing]

Homework Statement

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?

Homework Equations

Decay constant λ= 0.693 / T½
Where N = amount of radioactive substance,
N=N₀e^-λt

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

Remember, ln (e^x) = x

 

[InsaneScientist]

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

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.

log_eN = -λt
log_e0.35 = -1.209x10^-4t

t = log_e0.35 / -1.209x10^-4 = 8683.4 years.
Is this correct?

 

[SteamKing]

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.

log_eN = -λt
log_e0.35 = -1.209x10^-4t

t = log_e0.35 / -1.209x10^-4 = 8683.4 years.
Is this correct?

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

 

[SammyS]

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.

log_eN = -λt
log_e0.35 = -1.209x10^-4t

t = log_e0.35 / -1.209x10^-4 = 8683.4 years.
Is this correct?

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.

Spoiler

8678.5 years