Solveeit Logo

Question

Question: The decay constant of a radioactive isotope is \(\lambda\) if \(A_{1}\) and \(A_{2}\) are its activi...

The decay constant of a radioactive isotope is λ\lambda if A1A_{1} and A2A_{2} are its activities at times t1t_{1} and t2t_{2} respectively then the number of nuclei which have decayed during the time (t1t2)\left( t_{1} - t_{2} \right)

A

A1t1A2t2A_{1}t_{1} - A_{2}t_{2}

B

A1A2A_{1} - A_{2}

C

(A1A2)/λ\left( A_{1} - A_{2} \right)/\lambda

D

λ(A1A2)\lambda\left( A_{1} - A_{2} \right)

Answer

(A1A2)/λ\left( A_{1} - A_{2} \right)/\lambda

Explanation

Solution

: A1=λN1A_{1} = \lambda N_{1}at time t1t_{1}

A2=λN2A_{2} = \lambda N_{2}at time t2t_{2}

Therefore, number of nuclei decayed during time interval (t1t2)(t_{1} - t_{2}) is

N1N2=(A1A2)λN_{1} - N_{2} = \frac{(A_{1} - A_{2})}{\lambda}