The isotope of plutonium 238Pu is used to make thermoelectric power sources for spacecraft. Suppose that a space probe was launched in 2012 with 4.0 kg of 238Pu. Required:
a. If the half-life of 238Pu is 87.7 yr, write a function of the form Q(t)= Q0e- kt. to model the quantity Q(t) of 238Pu left after t-years.
b. If 1.6 kg of 238Pu is required to power the spacecraft's data transmitter, for how long will scientists be able to receive data?

A) Q(t) = 4e^-(0.0079t)

B) t = 115.99 ≈ 116

Therefore scientist will be able to receive data after 116 years

Step-by-step explanation:

a)

to write a function of the form Q(t)= Q₀e⁻^kt to model the quantity Q(t) of ²³⁸Pu left after t-years.

so given that; half-life of ²³⁸Pu is 87.7 years,

∴ t =  87.7 years , Q(t) = 0.5Q₀

Now we substitute these value in the form  Q(t)= Q₀e⁻^kt

Q(t)= Q₀e⁻^kt

0.5Q₀ = Q₀e^ -(87.7k)

0.5 = e^ -(87.7k)

now we take the natural logarithm of both sides

In(0.5) = Ine^ -(87.7k)

Now using the property logₙnᵃ = a

-87.7k = In(0.5)

k = - In(0.5) / 87.7

k = 0.0079

ALSO it was given that Q₀ = 4.0 kg

Therefore , model quality Q(t) of ²³⁸pu left after t years is:

Q(t) = 4e^-(0.0079t)

b)

to find the time left after 1.6kg of ²³⁸pu

we simple substitute  Q(t) = 1.6 into Q(t) = 4e^-(0.0079t)

so we have

1.6 = 4e^-(0.0079t)

e^-(0.0079t) = 1.6/4

e^-(0.0079t) = 0.4

again we take the natural logarithm of both sides,

Ine^-(0.0079t) = In(0.4)

again using the property logₙnᵃ = a

-0.0079t = In(0.4)

t = - in(0.4) / 0.0079

t = 115.99 ≈ 116

Therefore scientist will be able to receive data after 116 years

a = 3.14

b = 18

c = 18/5 or 3.6 or 3 3/5

step-by-step explanation: