Question #70676

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Answer 1 · 2017-04-16T13:42:19

The mass left is $= 6.39 g$ $=6.39g$

The half life of Cobalt 60 is $t_{\frac{1}{2}} = 5.27 y e a r s$ $t_(1/2)=5.27 years$

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

So,

$λ = \frac{ln 2}{5.27} = \frac{0.69}{5.27}$ $lambda=ln2/(5.27)=0.69/(5.27 )$

$= 0.1309 (y e a r s^{- 1})$ $= 0.1309(years^-1)$

We apply the equation

$A = A_{0} \cdot e^{- λ t}$ $A=A_0*e^(-lamdat)$

The activity is proportional to the mass.

$m = m_{0} \cdot e^{- λ t}$ $m=m_0*e^(-lamdat)$

$m = 200 \cdot e^{- (0.1309 \cdot 26.3)}$ $m=200*e^-(0.1309*26.3)$

$m = 200 \cdot e^{- 3.443}$ $m=200*e^-3.443$

$m = 200 \cdot 0.03195$ $m=200*0.03195$

$m = 6.391 g$ $m=6.391g$