Find N for Isotope with 10.84 Day Half-Life

[dinospamoni]

Homework Statement

The half life of a particular isotope is 10.84 days. Find the number of atoms of this isotope that would be necessary to produce a sample with an activity of 1.79 micro-Ci.

Homework Equations

R=N_{0} \lambda e^{-\lambda t}

where R is the decay rate
lambda is the decay constant
N is the number of atoms
and t is the half life

The Attempt at a Solution

First I converted the half life to seconds: 936576 s

than I solved for the decay rate constant: lambda= 7.401*10^-7 1/s

and R is 1.79*10^-6 Ci = 1.79*10^-6 decays/second

When I plugged these values into the equation though, I get 4.84, which is wrong.

 

[SammyS]

Homework Statement

The half life of a particular isotope is 10.84 days. Find the number of atoms of this isotope that would be necessary to produce a sample with an activity of 1.79 micro-Ci.

Homework Equations

R=N_{0} \lambda e^{-\lambda t}

where R is the decay rate
lambda is the decay constant
N is the number of atoms
and t is the half life

The Attempt at a Solution

First I converted the half life to seconds: 936576 s

than I solved for the decay rate constant: lambda= 7.401*10^-7 1/s

and R is 1.79*10^-6 Ci = 1.79*10^-6 decays/second

When I plugged these values into the equation though, I get 4.84, which is wrong.

Which equation did you plug these values into?

What did you use for t ?

 

[dinospamoni]

I'm not sure how I messed that up, but I totally didn't use that equation.

What I actually did was solve for lambda and then used the equation

N_{0} = \frac{R_0}{\lambda}

and got N to be 2.418, but that's wrong

 

[barryj]

According to Wikipedia,

"The Curie (symbol Ci) is a non-SI unit of radioactivity, named after Marie and Pierre Curie.[1][2] It is defined as

1 Ci = 3.7 × 10^10 decays per second. "

1 micro-Ci = 3.7E4 decays /sec so R should be 1.79 X 3.7E4 = 6.62 E4.

Maybe this is the problem

 

[barryj]

Deleted post.

It gave the answer to a Homework problem.

 

[dinospamoni]

That is definitely the problem. Thanks a ton!