What is the half-life in minutes of a radioactive nuclide if 1 ⁄ 32nd of the?

Question

What is the half-life in minutes of a radioactive nuclide if 1 ⁄ 32nd of the original sample remains after 18.0 min ?

Enter a numeric answer only do not include units

Answer 1

32 = 2^5, so the sample has already had five half-lives. Therefore the half-life must be 18/5 = 3.6 minutes.

3.6.

Answer 2

Half-life equation is as follows:

LN ( Nt / No ) = - k * t

Variable definitions:

Nt - number remaining after the time interval
No - initial number of nuclei (at time zero)
k - decay constant
t - time interval of decay

Because we do not have the decay constant, we need to use the following equation in order to find it:

k = LN(2) / x

x is the half life (commonly denoted as "t^(1/2)")

Applying this to your question, we end up with the following equation:

LN( [1/32] / 1 ) = - ( LN(2) / x ) * 18.0

Solving for x we obtain 3.6 minutes as a half-life.

Answer 3

3.60