$20$ % of a radioactive element disintegrates in $1 h r$ .The percentage of the radioactive element disintegrated in $2 h r s$ will be

Question

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Answer 1

Let initially the number of radioactive particles be $100$ .
Using the formula:
$N = N_{0} e^{- λ t}$ where $λ = \frac{2 . 3 0 3}{t} l o g (\frac{N _{0}}{N})$

$λ = \frac{2 . 3 0 3}{1} l o g (\frac{1 0 0}{8 0}) \Rightarrow λ = 0.223 h r s^{- 1}$

Radioactive material left in $2 h r s$ $= N_{0} e^{- λ t}$
$= 100 e^{- 0.223 \times 2}$
$= 63.99 \approx 64 %$

Radioactive material decayed in 2 hours $= (100 - 64) %$
$= 36$ %

Answer 2

Let initially the number of radioactive particles be

100

.
Using the formula:

N = N_{0} e^{- λ t}

where

λ = \frac{2 . 3 0 3}{t} l o g (\frac{N _{0}}{N})

λ = \frac{2 . 3 0 3}{1} l o g (\frac{1 0 0}{8 0}) \Rightarrow λ = 0.223 h r s^{- 1}

Radioactive material left in

2 h r s

= N_{0} e^{- λ t}

= 100 e^{- 0.223 \times 2}

= 63.99 \approx 64 %

Radioactive material decayed in 2 hours

= (100 - 64) %

= 36

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