Question

a 40 gram sample of a substance that's used for drug research has a k-value of 0.1446. find the substance's half life in days

Answer 1

Answer:

4.7935 days

Step-by-step explanation:

The expression that describes exponential decay is the following:

N(t) = No * $e^{-kt}$

Where N(t) is the mass of the substance after a period t of time.

No is the original amount of substance

k is the relative decay rate and t is the period of time elapsed.

The half life is the period of time after which the amount of substance has decreased by half.

We can issolate t in the expression, using the properties of logarithms:

$\frac{N(t)}{No}$=$e^{-kt}$

ln($\frac{N(t)}{No}$) = -k*t

t = - ln($\frac{N(t)}{No}$)/k = - ln($\frac{1}{2}$)/0.1446 = 4.7935 days.

Answer 2

The applicable decay formula is:N = No*e^(-kt)WhereN = Mass left after time tNo = Original massk = constantt = half lifeUsing the values given,N/No = e^(-0.1446t) = 1/2ln (1/2) = -0.1446 t *ln e^1-0.6931 = -0.1446 tt = 0.6331/0.1446 = 4.79 daysTherefore, half-life is 4.8 days.