Question

A radioactive sample has half life of 20 seconds. After 60 seconds of formation of sample, its activity is found to be $6.93\times {10}^{20}$ dps. During this time total energy released is ${10}^8$ J $(ln2=0.693)$

• A. Initial number of active atoms in sample is $1.6\times {10}^{23}$
• B. Number of active sample as t = 60 see is $1.2\times {10}^{23}$
• C. Energy released per fission is of order ${10}^{-15}$ J
• D. Initial number of active atoms is $2.4\times {10}^{21}$

Answer 1

Answer: (A), (C)

The correct options are
A Initial number of active atoms in sample is $1.6\times {10}^{23}$
C Energy released per fission is of order ${10}^{-15}$ J

Activity $A=$$\lambda N=N_o\lambda e^{-\lambda t}=\frac{N_o\lambda }{e^{\lambda t}}$
$N_o=\frac{A{e}^{\lambda t}}{\lambda }=\frac{A}{\lambda }{e}^{\frac{Ln2}{T}t}=\frac{A}{\lambda }{e}^{3\ln 2}=\frac{8A}{\lambda }$
$N_o=\frac{8\times 6.93\times {10}^{20}}{\ln \left(2\right)}\times 20=\frac{160\times \text{/}0.693{10}^{21}}{\text{/}0.693}$

$N_o=1.6\times {10}^{23}$
$N=N_o{e}^{-\lambda t}\Rightarrow N=\frac{N_o}{e^{\frac{ln2}{T}\times t}}=\frac{N_o}{8}$
$N=$$0.2\times {10}^{23}$

$\Delta N=N_o-N=1.4\times {10}^{23}$

Say energy per fission is $E$, then
$E\times 1.067\times {10}^{23}={10}^8\Rightarrow E\simeq {10}^{-15}J$