The half-life of the isotope Osmium-183 is 12 hours. Choose the equation below that gives the remaining mass of Osmium-183 in grams, M. after n half-lives have elapsed if there was an initial mass of 590 grams before decay. Then, use the equation to determine the mass remaining
after 36 hours have passed
Mn = 590. ()": M3 – 148
D
M, = 590 C4)*-* : My ~ 148
Mn = 590 - ()": Mz - 74
Mi = 590 - (1) * + : Mz - 74

The given equations are incomprehensible, I'm afraid...

You're given that osmium-183 has a half-life of 12 hours, so for some initial mass M₀, the mass after 12 hours is half that:

1/2 M₀ = M₀ exp(12k)

for some decay constant k. Solve for this k :

1/2 = exp(12k)

ln(1/2) = 12k

k = 1/12 ln(1/2) = - ln(2)/12

Now for some starting mass M₀, the mass M remaining after time t is given by

M = M₀ exp(kt )

So if M₀ = 590 g and t = 36 h, plugging these into the equation with the previously determined value of k gives

M = 590 exp(36k) = 73.75

so 73.75 ≈ 74 g of Os-183 are left.

Alternatively, notice that the given time period of 36 hours is simply 3 times the half-life of 12 hours, so 1/2³ = 1/8 of the starting amount of Os-183 is left:

590/8 = 73.75 ≈ 74

sorry not sure

step-by-step explanation:

other answer is correct. i had used the radius of the radius somehow. a website: