Help :(( Exponential Decay

Submitted by Aesha on

a radio active element follows the law of exponential change and has a half-life of 38 hours. how long will it take for 90% of the radioactivity of the element to be dissipated ?

Jhun Vert Fri, 08/28/2015 - 07:13

Following exponential change
$x = x_o e^{-kt}$
 

After 38 hours, 50% is dissipated, hence 50% retained
$0.5x_o = x_o e^{-38k}$

$0.5 = e^{-38k}$

$e^{-k} = 0.5^{1/38}$
 

Thus,
$x = x_o 0.5^{t/38}$

90% dissipated means 10% retained, hence, x = 0.10xo
$0.10x_o = x_o 0.5^{t/38}$

$0.10 = 0.5^{t/38}$

$t = 126.23 ~ \text{hrs}$

Jhun Vert Fri, 08/28/2015 - 21:46

You can also use exponential STAT in your calculator:

CASIO fx-991ES PLUS: [MODE]   [3:Stat]   [5:e^X]

  X (time) Y (% retained)
1 0 1
2 38 0.5

[AC]
For 10% retained:
X (time) = 0.1x-caret = 126.23 hours

x-caret can be found at [Shift]   [1:Stat]   [5:Reg]   [4:x-caret]