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The activity of a radioactive substance ...

The activity of a radioactive substance is 4700 per minute. Five minute later the activity is 2700 per minute. Find
(a) decay constant and
(b) half-life of the radioactive substance.

Text Solution

Verified by Experts

We have `A= A_0 e^(- lamda t)`
` (A)/(A_0) = e^(- lamda t)`
`In [(A)/(A_0)]= - lamda t`
` In [(A_0)/(A)]= lamda t`
` (2.303 )/(t) log _(10) [(A_0)/(A) ] = lamda`
Here `A_0 = 4700 ` per minute
`A = 2700 ` per minute
` A= 2700 ` per iminute & t=5 minute
`(2.303 )/(5) log (1.7407 )`
`=0.4606 xx 0.2407 `
` lamda = 0.11086 ` per minute
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