Question

Half life of a substance is 20 minutes, then what is the time between 33°/• decay and 67°/• decay?

Answer 1

For 67% decay,
Remaining = 100% - 67% = 33% = 0.33
0.33 = $e^{-t1. ln(2)/20}$
⇒ ln(0.33) = -t1. ln(2) / 20
⇒t1 = -[ ln(0.33)×20÷ln(2) ]
⇒t1 = 31.99 min

Similarly
For 33% decay,
Remaining = 100% - 33% = 67% = 0.67
0.67 = $e^{-t2. ln(2)/20}$
⇒t2 = -[ ln(0.67)×20÷ln(2) ]
⇒t2 = 11.55 min

Difference = t1 - t2 = 20.44 minutes

Answer 2

Answer:T1/2 = 20min

N1/N0 = 67/100 & N2/N0 = 33/100

°•°N/N0 = (1/2)^t/T

67/100 = (1/2)^t1/20. ....(1)

33/100 = (1/2)^t2/20. ....(2)

From (1) & (2) :-

(33/100)*(100/67)= (1/2)^t2/20/(1/2)^t1/20

(1/2)^1 = (1/2)^(t2-t1)/20

Hence: t2-t1= 20mins.

;)

Explanation: Source: Ncert book XIIth(modern physics) Hit thanks if you like;)

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