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andrew11 [14]
3 years ago
8

The half-life of the isotope Osmium-183 is 12 hours. Choose the equation below that gives the remaining mass of Osmium-183 in gr

ams, 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
Mathematics
1 answer:
raketka [301]3 years ago
6 0

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 <em>M</em>₀, the mass after 12 hours is half that:

1/2 <em>M</em>₀ = <em>M</em>₀ exp(12<em>k</em>)

for some decay constant <em>k</em>. Solve for this <em>k</em> :

1/2 = exp(12<em>k</em>)

ln(1/2) = 12<em>k</em>

<em>k</em> = 1/12 ln(1/2) = - ln(2)/12

Now for some starting mass <em>M</em>₀, the mass <em>M</em> remaining after time <em>t</em> is given by

<em>M</em> = <em>M</em>₀ exp(<em>kt </em>)

So if <em>M</em>₀ = 590 g and <em>t</em> = 36 h, plugging these into the equation with the previously determined value of <em>k</em> gives

<em>M</em> = 590 exp(36<em>k</em>) = 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

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