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 ?

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 = 0.10x_{o} $0.10x_o = x_o 0.5^{t/38}$

$0.10 = 0.5^{t/38}$

$t = 126.23 ~ \text{hrs}$

You can also use exponential STAT in your calculator:

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

[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]

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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 = 0.10x

_{o}$0.10x_o = x_o 0.5^{t/38}$

$0.10 = 0.5^{t/38}$

$t = 126.23 ~ \text{hrs}$

You can also use exponential STAT in your calculator:

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

X(time)Y(% retained)[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]

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