Question

How much of an 800-gram sample of potassium-40 will remain after 3.9 x 109 years of radioactive decay?

Answer 1

$N_{t} = N_{o} e^{ \frac{-693t}{ t_{ \frac{1}{2} } } }$

Where,
$N_{t} = Quantity remaining after some time$
$N_{o} = Original qunatity$
$t_{ \frac{1}{2} } = Half life = 1.251* 10^{9} years for potassium-40$

Substituting;
$N_{} =800* e^{ \frac{-0.693*3.9* 10^{9} }{1.251* 10^{9} } }$ = 151.83 g

Answer 2

Ans: Amount of K-40 remaining = 92.2 g

Given:

Initial mass of potassium-40 = M₀ = 800 g

Time, t = 3.9 *10⁹ years

To determine:

The mass of potassium-40 (M(t)) that will remain after time 't'

Explanation:

The radioactive decay equation is given as:

$\frac{M(t)}{M_{0} } = e^{\frac{-0.693t}{t_{1/2} } }$ ----(1)

where t1/2 is the half-life of K-40 = 1.251*10⁹ years

Substitute for M₀, t and t1/2 in equation(1):

$M(t) = 800 e^{\frac{-0.693*3.9*10^{9} }{1.251*10^{9} } } = 92.2 grams$