The activity of a radioactive sample is 64 ti | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The activity of a radioactive sample at time $t$ is given by the formula: $A = A_0 \left( \frac{1}{2} ight)^{t / T_{1/2}}$.Given: $A_0 = 64 	im

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(D) The activity of a radioactive sample at time $t$ is given by the formula: $A = A_0 \left( \frac{1}{2} ight)^{t / T_{1/2}}$.Given: $A_0 = 64 	imes 10^{-5} 	ext{ Ci}$,$A = 5 	imes 10^{-6} 	ext{ Ci}$,and $T_{1/2} = 3 	ext{ days}$.Substituting the values: $5 	imes 10^{-6} = 64 	imes 10^{-5} \left( \frac{1}{2} ight)^{t / 3}$.Dividing both sides by $64 	imes 10^{-5}$: $\frac{5 	imes 10^{-6}}{64 	imes 10^{-5}} = \left( \frac{1}{2} ight)^{t / 3}$.$\frac{0.5}{64} = \left( \frac{1}{2} ight)^{t / 3} \Rightarrow \frac{1}{128} = \left( \frac{1}{2} ight)^{t / 3}$.Since $\frac{1}{128} = \left( \frac{1}{2} ight)^7$,we have $\left( \frac{1}{2} ight)^7 = \left( \frac{1}{2} ight)^{t / 3}$.Equating the exponents: $7 = \frac{t}{3} \Rightarrow t = 21 	ext{ days}$.

The activity of a radioactive sample is $64 \times 10^{-5} \text{ Ci}$. Its half-life is $3 \text{ days}$. The activity will become $5 \times 10^{-6} \text{ Ci}$ after how many days?

Similar Questions

The activity of a radioactive element decreases to one-third of its original activity $R_0$ in $9$ years. After a further $9$ years,its activity will be

$A$ count rate meter shows a count of $240$ per minute from a given radioactive source. One hour later,the meter shows a count rate of $30$ per minute. The half-life of the source is .......... $min$.

$A$ radioactive substance has a half-life of $1$ year. The fraction of this material that would remain after $5$ years will be

The activity of a sample is $64 \times 10^{-5} \, Ci$. Its half-life is $3 \, days$. The activity will become $5 \times 10^{-6} \, Ci$ after ......... $days$.

The half-life of a radioactive element which has only $\frac{1}{32}$ of its original mass left after a lapse of $60\, days$ is ........ $days$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module