Calculating Half-life of 6He through Beta Minus Decay

[physicsstudent14]

Homework Statement

A student takes a sample of 32,000 6He atoms and measures 31000 daughter nuclei 4 seconds later. What is the half-life of 6He? What is the daughter nucleus if the 6He underwent beta minus decay?

Homework Equations

$$N = (No)(0.5)^{t/halflife}$$

The Attempt at a Solution

No = 32000 atoms = 32000 nuclei
N = 31000 nuclei
1 atom = 1 nuclei
t = 4 s
Find halflife

$$N/No = (0.5)^{t/halflife}$$


$$log(N/No)/log(0.5) = t/halflife$$

$$log(N/No)/log(0.5) = t/halflife$$

$$halflife = t/log(N/No)/log(0.5)$$

$$halflife = t(log(0.5))/log(N/No)$$

$$halflife = (4 s)(log 0.5)/log(31000/32000)$$

I get half-life = 87.3 seconds, but the real half-life of 6He is about 806.7 ms. Does my answer and problem solving method look ok? Also, I think the daughter nucleus is 6Li because in beta minus decay causes the atomic number of the atom to go up by one as the neutron becomes a proton and an emitted electron. Does this make sense? I'm sorry for asking silly, easy questions, but we have not covered this material in class, so I am relying on my book to answer these questions. Thank you so much for your help.

 

[hage567]

If you start with 32000 6He, and see 31000 daughter nuclei, what does that tell you about the change in the number of 6He nuclei? No is the # of 6He you started with, therefore N must be the # of 6He you have left after the 4 seconds. The number of 6He nuclei left is not 31000!

 

[physicsstudent14]

Oh...now I understand, I hope.

No = 32000 atoms = 32000 nuclei
N = 32000-31000 = 1000 nuclei
1 atom = 1 nuclei
t = 4 s
Find halflife

$$N/No = (0.5)^{t/halflife}$$


$$log(N/No)/log(0.5) = t/halflife$$

$$log(N/No)/log(0.5) = t/halflife$$

$$halflife = t/log(N/No)/log(0.5)$$

$$halflife = t(log(0.5))/log(N/No)$$

$$halflife = (4 s)(log 0.5)/log(1000/32000)$$

Now this gives me half-life = 0.8 seconds, which is closer to the accepted value of 806.7 ms. I hope this is right. Thanks!