# The Half-life of a Radioisotope is 10 H. Find the Total Number of Disintegration in the Tenth Hour Measured from a Time When the Activity Was 1 Ci. - Physics

Sum

The half-life of a radioisotope is 10 h. Find the total number of disintegration in the tenth hour measured from a time when the activity was 1 Ci.

#### Solution

Given:
Half-life of radioisotope,T_(1"/"2) = 10 hrs

Initial activity, A_0 = 1 Ci

Disintegration constant, lambda = 0.693/(10 xx 3600)  "s"^-1

A = A_0e^(-lambdat)

Here, A0 = Initial activity
lambda = Disintegration constant
t = Time taken

After 9 hours,

Activity , A = A_0e^(-lambdat) = 1 xx e^(-0.693/(10 xx 3600) xx 9 = 0.536 Ci

therefore Number of Atoms left , N = A/lambda = (0.536 xx 10 xx 3.7 xx 10^10 xx 3600)/0.693 = 103.023 xx 10^13 After 10 hrs

Activity , A " = A_0e^(-lambdat)

= 1 xx e^(-0.693/10 xx 10) = 0.5 Ci

Number of atoms left after the 10th hour (N ")will be

A" = lambdaN "

N" = A"/lambda

= (0.5 xx 3.7 xx 10^10 xx 3.600)/(0.693"/"10)

= 26.37 xx 10^10 xx 3600 = 96.103 xx 10^13

Number of disintegrations  = (103.023 − 96.103) × 1013
= 6.92 × 1013

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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 24 The Nucleus
Q 26 | Page 443