1. May 6, 2008

### shigg927

1. The problem statement, all variables and given/known data
Thorium (with half-life T1/2 = 1.913 yr. and atomic mass 228.028715 u) undergoes alpha decay and produces radium (atomic mass 224.020186 u) as a daughter nucleus. (Assume the alpha particle has atomic mass 4.002603 u.)

What percent of thorium is left after 266 days?

2. Relevant equations

X --> Y + He
N=No*(1/2)^n
n= t/T(half)
T(half)= .693/$$\lambda$$

3. The attempt at a solution

I found that lambda=4.14x10^-5 hrs^-1 (the problem asks for it in hours, dumb, I know.)

I then found the number of half-lives to be 266 days, or 6384 hours divided by 16757.88 hours, to be .381 half-lives. I multiplied this by Thorium's atomic mass to get 36% but this keeps turning up incorrect for my online homework.

2. May 7, 2008

### kamerling

Why don't you just apply N=No*(1/2)^n ?

since 266 days is shorter than the half life, more than 50% should be left.

3. May 7, 2008

### shigg927

Ahhhhh for some reason I thought I needed to now the number of nuclei, did NOT know I could just use the atomic mass. I got it, thank you!