A certain radioactive element decays over time according to the equation y=A(1/100)^t/300 where A= the number of grams present initially and t=time in years. If 1000 grams were present initially, how many grams will remain after 900 years?

Question

A certain radioactive element decays over time according to the equation y=A(1/100)^t/300 where A= the number of grams present initially and t=time in years.  If 1000 grams were present initially, how many grams will remain after 900 years?

anonymous · Answer

please help(:

anonymous · Answer

you are aware these equations are type first order reactions, right?

anonymous · Answer

wait what?

anonymous · Answer

Chemical kinetics. NVM, we'll simplify the equation before solving it. Take log on both sides.

anonymous · Answer

how would I set it up exactly?

anonymous · Answer

that gives us lig (y) = (t/300)*log(1/100) + log A

anonymous · Answer

which gives log y= log A- 2(t/300)

anonymous · Answer

differenciating w.r.t. T gives us the rate of decay.

anonymous · Answer

wouldnt't I just plug in the values to the equation? like a and t?

anonymous · Answer

the value of y in the eqn gives ammount remaining and the ammount decayed would be by subtracting from the inintial..

anonymous · Answer

confused

anonymous · Answer

sorry, I thought you needed to take the rate of reaction. =P

Just put the values there and solve. That gives you the value required.

anonymous · Answer

1000 (1/100)^900/300
=1000*(1/100)^3
=0.001 is what is left.

anonymous · Answer

thats what I got and its not one of the answer choices... here ill post the choices

anonymous · Answer

$$y=A\left(1\over100\right)^{t\over300}$$
plug in A = 1000
and t = 900

anonymous · Answer

these are the choices
A) 500 grams
B) 250 grams
C)125 grams
D) 62.5 grams

anonymous · Answer

@sparky16 is that how the equation looks? if not, use the equation editor below to enter it clearly.

anonymous · Answer

oh wait I typed it wrong...lemme fix it in the equation editor

anonymous · Answer

$$y=A(\frac{ 1 }{ 2 })^{\frac{ t }{ 100 }}$$

anonymous · Answer

100(1/2)9
which is 25/32

anonymous · Answer

pardon 125/16.

anonymous · Answer

@sparky16 I still think you've got it wrong

anonymous · Answer

is it perhaps like this:
$$y=A\left(1\over2\right)^{t\over300}$$

anonymous · Answer

1.953125

anonymous · Answer

$$y=A\left(1\over2\right)^{\Large\color{red}{t\over300}}$$

anonymous · Answer

yes. it is.. my bad.

anonymous · Answer

ok, now plug in the numbers for "A" and "t"

anonymous · Answer

125?

anonymous · Answer

good.

anonymous · Answer

be careful when you note down the equation.

anonymous · Answer

@NeetziD do not get disappointed. you would have got it right, if the equation was not mis-leading.

anonymous · Answer

No, I'm not. Thanks, by the way. I didn't read the requirement, and since chemistry is too high... been taking it in that direction. =P