Exponential Decay?

Question

Bismuth-210 has a half life of 5 days. use the formula m(t) = me^kt
a. a sample originally has a mass of 800mg. Find a formula for the mass remaining after t days
b. find the mass remaining after 30 days
c. when is the mass reduced to 1mg?

Answer 1

Hi,

a. a sample originally has a mass of 800mg. Find a formula for the mass remaining after t days

A = 800e^(-.13863t)

b. find the mass remaining after 30 days

A = 800e^(-.13863t) if t = 30 is:
A = 800e^(-.13863*30)
A = 12.5

c. when is the mass reduced to 1mg?

A = 800e^(-.13863t) if A = 1
1 = 800e^(-.13863t)
.00125 = e^(-.13863t)
ln .00125 = ln e^(-.13863t)
ln .00125 = -.13863t
ln (.00125)/(-.13863) = t
48.219 = t so it will take 48.2 days for the mass to reduce to 1mg.

I hope those help!! :-)

Answer 2

Part a
m( t ) = m e^( k t )
m / 2 = m e^(5 k)
1/2 = e^(5 k)
ln(1/2) = 5 k
k = (1/5) ln (1/2)
k = - 0.139
m(t) = 800 e^(-0.139 t)

Part b
m(30) = 800 e^(- 0.139 x 30)
m(30) = 12.37 mg

Part c
1 = 800 e^(-0.139 t)
e^(0.139 t) = 800
0.139 t = ln 800
t = ln 800 / 0.139
t = 48.1 days

Answer 3

a. m = m0 e^(-kt). Since you are told m0 = 800 mg and k = 1/5 (days)^-1

m = 800 e^(-t/5) mg t is in days.

b. m = 800 e^(-30/5) =800 e^(-6) = 1.98 mg

c: Start with m = m0 e^(-kt) solve for t

e^(-kt) = m/m0

-kt = ln(m/m0) ln = natural log

t = -1/k ln(m/m0)

now let m =1 mg ---> t = -5*ln(1/800) = 33.42 days

Answer 4

you have to find out what k is first:

a) m(t)=me^kt
well after 5 days you would have 400mg so:
m(5)=(800mg)e^(k*(5days))=400mg

now solve for k:

e^5k=(400mg)/(800mg)=0.5
Ln(e^5k)=Ln(0.5) so 5k=-0.69, therefore k=-0.139 (1/days)

b) m(30)=(800mg)e^(-0.139*30) = 12.5mg

c) 1mg = (800mg)e^(-0.139*t)

Ln(1mg/800mg) = Ln(e^-0.139*t)

-6.685 = -0.139*t therefore t = 48 days