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(C) The radioactive decay law is given by $A = A_0 	imes (1/2)^{t/T}$,where $A$ is the activity at time $t$,$A_0$ is the initial activity,and $T$ is

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$15$

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(C) The radioactive decay law is given by $A = A_0 	imes (1/2)^{t/T}$,where $A$ is the activity at time $t$,$A_0$ is the initial activity,and $T$ is the half-life.Given,$A_0 = 64 	imes A_{safe}$ and we want to find the time $t$ when $A = A_{safe}$.Substituting these values: $A_{safe} = 64 	imes A_{safe} 	imes (1/2)^{t/T}$.This simplifies to $1/64 = (1/2)^{t/T}$.Since $64 = 2^6$,we have $(1/2)^6 = (1/2)^{t/T}$.Therefore,$t/T = 6$,which means $t = 6 	imes T$.Given the half-life $T = 2$ hours $30$ minutes = $2.5$ hours.Thus,$t = 6 	imes 2.5 = 15$ hours.

$A$ freshly prepared radioactive source of half-life $2$ hours $30$ minutes emits radiation which is $64$ times the permissible safe level. The minimum time,after which it would be possible to work safely with the source,will be in hours.

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