Please help with my home work

Question

The half-life of Palladium-100 is 4 days. After 20 days a sample of Palladium-100 has been reduced to a mass of 2 mg.

1. What was the initial mass (in mg) of the sample
2. What is the mass (in mg) 6 weeks after the start?
3. You may enter the exact value or round to 4 decimal places.

Answer 1

half life = 4 days

after 20 days 2 miligrams are left.

what was the original amount on day 0?

day 20, 2 mg

day 16, 4 mg

day 12, 8 mg

day 8, 16 mg

day 4, 32 mg

day 0, 64 mg

day x, (2)^((20+4-x)/4)

day 20, 2^((24-20)/4)=2

day 0, (2)^((24-0)/4)=2^6=64

day 42, 2^((24-42)/4)=2^(-18/4)=2^-4.5=1/2^4.5=

in 6 weeks = 6(7) = 42 days, 1/16 mg is left

20 days 2 mg

24 days 1 mg

28 days 1/2 mg = 0.5 mg

32 1/4 mg = 0.25 mg

36 1/8 = 0.125 mg

40 days 1/16 mg = 0.0625 mg

42 days, roughly half way between 0.6 and .03 = .045

44 day 1/32= 0.03125 mg

A = Pe^rt is the general formula for growth or decay

where t = time, P=original amount, A= ending amount and r=the growth or decay rate,

it's the same formula for continuously compounded interest

if r>0, it's a growth rate, if r<0, it's a decay rate

1/2 = e^4r with t=4 days, for half life

ln.5 = 4r

r = ln.5/4 = about -0.1733 = -17.33% per day decay rate

for 2 grams after 20 days:

2 = Pe^20r = Pe^20(-.1733)

2/P = e^20(-.1733)

2/P =e^-3.466

P = 2/e^-3.466 = 2e^3.466 = 64 mg = initial amount

after 6 weeks=6(7) days = 42 days

Amount remaining =

A =exactly 64e^(42(ln.5)/4))

= approximately 64e^42(-.1733)

= 64e^-7.2786

=64/1448.96=0.0442

or

Amount left after 6 weeks or 42 days =

2^-4.5 =approximately 0.0442