Question

Francium has a decay rate of 50% every 22 minutes. How much of an initial amount of 4.8 grams would be left after 88 minutes?
t=88/22=4A=Aoe^-kt

Answer 1

Answer:

0.65g

Step-by-step explanation:

Decay rate = 50%

Decay rate = 0.5

$A = Aoe {}^{ - kt}$

where Ao = initial amount at start, k = rate of decay or growth, t = time, e = euler's number and A = value at time 't'

Ao = 4.8

t = 88/44 = 4

k = 0.5

substitute these values in the equation

$A = 4.8e {}^{ - 0.5 \times 4}$

$A = 0.6496$

$A = 0.65 \: (2dp)$