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Question: The mean lives of a radioactive substance are 1620 year and 405 year for a-emission and b-emission r...

The mean lives of a radioactive substance are 1620 year and 405 year for a-emission and b-emission respectively. Find the time during which three-fourth of a sample will decay if it is decaying both by a-emission and b-emission simultaneously.

A

249 years

B

449 years

C

133 years

D

99 years

Answer

449 years

Explanation

Solution

The decay constant l is the reciprocal of the mean life t. Thus, la = 11620\frac { 1 } { 1620 } per year and

lb = 1405\frac { 1 } { 405 }per year \Total decay constant, l = la + lb or

l = 11620+1405=1324\frac { 1 } { 1620 } + \frac { 1 } { 405 } = \frac { 1 } { 324 } per year

We know that N = N0 e–lt When 34\frac { 3 } { 4 } th part of the sample has disintegrated, N = N0/4

\ = N0e–lt or elt = 4 Taking logarithm of both sides, we get lt = loge 4 or t = 1λ\frac { 1 } { \lambda }loge 22 = 2λ\frac { 2 } { \lambda } loge 2

= 2 × 324 × 0.693 = 449 year