radioactive gallium-67 decays by 1.48% every hour; there are 100 milligrams initially.
a) find a formula for the amount of agallium-67 remaining after t hours.
b) how many milligrams are left after 24 hours? After 1 week?
2007-02-22 03:51:06 · 4 answers · asked by Cindie lou 1 in Science & Mathematics ➔ Mathematics
Lets try a formule like
G=k*e^bt
At t=0 G =100 = k*e^0=k. So k= 100
After 1 hour there are 98.52 mg so 98.52 = 100 e^b
so e^b= 98.52/100 and b= ln(0.9852) =-0.0149
G(in mg) = 100*e^(-0.0149t) After one week = 7*24 hours=168h
G=100*e^-(2.5032) = 8.17 mg
2007-02-22 04:15:15 · answer #1 · answered by santmann2002 7 · 0⤊ 0⤋
y = C e^-.0148t , where y = amount of gallium-67, t = time in hours and C = 100 mg.
So, y = 100e^-.0148t.
When t = 24, y = 100e^-1.48 = 22.764mg
When t = 1 week = 168 hours, y = 100e^-2.486 = 8.324mg
2007-02-22 12:07:31 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
a is amount
t is time in hours
a=100 x0.9852 ^t
so after 24 hours=69.91746537481 etc.
after a week=8.16769914 etc.
2007-02-22 12:05:48 · answer #3 · answered by thom 2 · 0⤊ 0⤋
(x*(1,0148)^t)-x
where x is in miligrams, you´ll find out what´s left in "t" hours
at least i think...
2007-02-22 12:03:53 · answer #4 · answered by Gabriel G 3 · 0⤊ 0⤋