Posted by amy on Thursday, February 3, 2011 at 12:03am.
The reactions involved are
140Ba -> 140La + e-,
half life 12.8 days
and
140La -> 140Ce(stable) + e-,
half life 40.2 hours
Set up and solve the differential equation for [140La] and solve for the time is reaches its maximum value, which would be when d/dt[La] = 0
The population of 140Ba is
[140Ba^]*(1/2)^(t/12.8)
This will allow you to write an equation for the rate La is created. Set that equal to the rate La is destroyed by decay. You should be able to solve for [La] at that time. It is probably easier to do with decay coefficients rather than half lives.
The reactions involved are
140Ba -> 140La + e-,
half life 12.8 days
and
140La -> 140Ce(stable) + e-,
half life 40.2 hours
Set up and solve the differential equation for [140La] and solve for the time is reaches its maximum value, which would be when d/dt[La] = 0
The population of 140Ba is
[140Ba^]*(1/2)^(t/12.8)
This will allow you to write an equation for the rate La is created. Set that equal to the rate La is destroyed by decay. You should be able to solve for [La] at that time. It is probably easier to do with decay coefficients rather than half lives.
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