M=Mo e^(-kt)?

Question

where k and Mo are constants.
intially, there are 93 grammes of the element and after 18.8 years, there are 57.9 grammes left.

1)Find the value of K in year^(-1)
2)Find the half life of the element

Answer 1

1)
M = Mo e^(-kt)
e^(-kt) = (M/Mo)
-kt = ln(M/Mo)
k = (-1/t)·ln(M/Mo)
= (-1/18.8)·ln(57.9/93) = 0.025 yr^-1

2)
M = Mo/2 = Mo e^(-kt) ──► ½ = e^(-kt)
-kt = ln(½)
t = (-1/k)·ln(½)
= (-1/0.025)·ln(½) = 27.5 yr

Answer 2

93 grammes when t=0 implies Mo = 93
57.9 g when t= 18.8 (if t is in years) if not convert to seconds if needed)
solve for k to get the constant.

2) sub in grammes = 46.5 (1/2 of 93) find t. this gives you the 1/2 life.

you will need to use logs!!

Answer 3

You write
57.9 = 93 e(-Kt ) hence 57.9 / 93 = e (-Kt) inverse the fraction

93/57.9 = e (Kt) ----Kt = ln (93/57.9)

T = ln (93/57.9)/18.8---> ln (1.606) = 0.473 =18.8*K

Answer K = 0.025 year^(-1)

2) Half life is the time necessary to have half of the atoms


replace m by Mo/2 ----¨> Mo/2 = Mo e^(-KT) WhereT is the half life. Divide by Mo

1/2 = e(-^KT) hence 2 = e^ (KT)

Ln 2 = KT and T = ln2/K (Remember this formula!)

here T = ln 2 / 0.025 = 27.72 years

Answer 4

well solve for k
(ln can be applied to both sides of the equation and cancels e)
k = -ln(M/Mo)/t
where M =57.9, Mo=93 t= 18.8
remember your units
Half life you can find easly yourself

Answer 5

93=57.9e^(-18.8k)
solve for k
once k trickles out
substituting in
1/2M0=M0e^(-kt)
t will give you the half lifeperiod