The radioactivity of a sample is A1 at time t1 and A2 at time t2 . If the mean life of the specimen is T, the number of atoms that have disintegrated in the time interval of (t2-t1) is

Question

The radioactivity of a sample is A1 at time t1 and A2 at time t2 . If the mean life of the specimen is T, the number of atoms that have disintegrated in the time interval of (t2-t1) is (A)  (A1-A2)  (B)  ((A1-A2)/T)  (C)  (A1-A2)T  (D)  (A1t1-A2t2)

Tardigrade Admin · Accepted Answer

A1=N1 λ, A2=N2 λ Mean life, T=(1/λ) A1-A2 =(N1-N2) λ =(N1-N2) (1/T) So, number of atoms disintegrated in (t2-t1) s =(N1-N2)=(A1-A2) T

Q. The radioactivity of a sample is $A_{1}$ at time $t_{1}$ and $A_{2}$ at time $t_{2}$ . If the mean life of the specimen is $T$ , the number of atoms that have disintegrated in the time interval of $(t_{2} - t_{1})$ is

$(A_{1} - A_{2})$

$\frac{( A _{1} - A _{2} )}{T}$

$(A_{1} - A_{2}) T$

$(A_{1} t_{1} - A_{2} t_{2})$

$A_{1} = N_{1} λ, A_{2} = N_{2} λ$
Mean life, $T = \frac{1}{λ}$
$A_{1} - A_{2} = (N_{1} - N_{2}) λ$
$= (N_{1} - N_{2}) \frac{1}{T}$
So, number of atoms disintegrated in $(t_{2} - t_{1}) s$
$= (N_{1} - N_{2}) = (A_{1} - A_{2}) T$

Q. The radioactivity of a sample is A1​ at time t1​ and A2​ at time t2​ . If the mean life of the specimen is T, the number of atoms that have disintegrated in the time interval of (t2​−t1​) is

(A1​−A2​)

T(A1​−A2​)​

(A1​−A2​)T

(A1​t1​−A2​t2​)