If half life of an element is 69.3 hours then how much of its percent will decay in 10^{text {th }}

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13.Nuclei

hard

If half life of an element is $69.3$ hours then how much of its percent will decay in $10^{\text {th }}$ to $11^{\text {th }}$ hours. Initial activity $=50\, \mu Ci$

A

$1$

B

$2$

C

$3$

D

$4$

(AIIMS-2019)

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 $=\left(1-\frac{1}{e^{0.01}}\right) \times 100$

$=1 \%$

Standard 12

Physics

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