- Home
- Standard 12
- Physics
$50\, \mu Ci$ શરૂઆતની એક્ટીવિટી ધરાવતા રેડિયોએક્ટિવ તત્વનો અર્ધઆયુ $69.3$ કલાક છે , તો $10^{\text {th }}$ અને $11^{\text {th }}$ વચ્ચે વિભંજન પાતનાં ન્યુક્લિયસની ટકાવારી શોધો.
$1$
$2$
$3$
$4$
Solution
Consider active Nuclei at $t =0$ is $N_{0}$ then,
The active nuclei at $t =10 hr$
$N_{1}=N_{0} e^{-10 \lambda}$
The active nuclei at $t=11 hr$
$N_{2}=N_{0} e^{-11 \lambda}$
The expression for the decay percentage,
$\%$ decay $=\left(\frac{N_{1}-N_{2}}{N_{1}}\right) \times 100$
$=\left(\frac{N_{0} e^{-10 \lambda}-N_{0} e^{-11 \lambda}}{N_{0} e^{-10 \lambda}}\right)$
$=\left(1-\frac{1}{e^{\lambda}}\right) \times 100$
$=\left(1-\frac{1}{e^{\ln 2 / T_{1 / 2}}}\right) \times 100 \times 100$
The half life of the element is, $T_{1 / 2}=69.3 hr$ therefore,
$\%$ decay $=\left(1-\frac{1}{e^{0.01}}\right) \times 100$
$=1 \%$
