The half-life of P-32 is 14.30 days. How many milligrams of a 20.00 mg sample of P-32 will remain after 85.80 days?

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The half-life of P-32 is 14.30 days. How many milligrams of a 20.00 mg sample of P-32 will remain after 85.80 days?

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Answer 1 · 2017-08-04T20:27:27

The amount remaining is $= 0.3125 m g$ $=0.3125mg$

Explanation:

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The half life is $t_{\frac{1}{2}} = 14.3 d$ $t_(1/2)=14.3d$

This means that after $14.3$ $14.3$ days $50 %$ $50%$ of the sample will remain

The time is $T = 85.8$ $T=85.8$ days

This is equal to $= \frac{85.8}{14.3} = 6$ $=85.8/14.3=6$ half lives

Therefore,

The amount remaining is $= \frac{1}{2^{6}} \cdot 20 = 0.3125 m g$ $=1/2^6*20=0.3125mg$

We can solve this problem with the equation

$m = m_{0} e^{- λ t}$ $m=m_0e^(-lambdat)$

The radioactive constant is $λ = \frac{ln 2}{t_{\frac{1}{2}}} = \frac{ln 2}{14.3}$ $lambda=ln2/t_(1/2)=ln2/14.3$

Therefore,

$m = 20 \cdot e^{- ln 2 \cdot \frac{85.3}{14.3}} = 20 \cdot e^{- 6 ln 2} = 20 \cdot e^{- ln (2^{6})} = 20 \cdot e^{- ln 64}$ $m=20*e^(-ln2*85.3/14.3)=20*e^(-6ln2)=20*e^-ln(2^6)=20*e^-ln64$

$m = \frac{20}{64} = 0.3125 m g$ $m=20/64=0.3125mg$

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The half-life of P-32 is 14.30 days. How many milligrams of a 20.00 mg sample of P-32 will remain after 85.80 days?

1 Answer

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