A radioactive sample decays frac{7}{8} times | vedclass

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

Question

(A) Let the initial quantity be $N_0$.Given that the sample decays by $\frac{7}{8}$ of its original quantity,the remaining quantity $N$ is:$N = N_0 -

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Let the initial quantity be $N_0$.Given that the sample decays by $\frac{7}{8}$ of its original quantity,the remaining quantity $N$ is:$N = N_0 - \frac{7}{8}N_0 = \frac{1}{8}N_0$.We know the radioactive decay law is $N = N_0 (\frac{1}{2})^n$,where $n$ is the number of half-lives.$\frac{1}{8}N_0 = N_0 (\frac{1}{2})^n$$(\frac{1}{2})^3 = (\frac{1}{2})^n$Thus,$n = 3$.Since $n = \frac{t}{T_{1/2}}$,where $t = 15$ minutes and $T_{1/2}$ is the half-life:$3 = \frac{15}{T_{1/2}}$$T_{1/2} = \frac{15}{3} = 5$ minutes.

$A$ radioactive sample decays $\frac{7}{8}$ times its original quantity in $15$ minutes. The half-life of the sample is $......$ minutes.

Similar Questions

Two radioactive materials $A$ and $B$ have decay constants $5\lambda$ and $\lambda$ respectively. At $t=0$,they have the same number of nuclei. The ratio of the number of nuclei of $A$ to that of $B$ will be $(1/e)^2$ after a time interval of:

The counting rate observed from a radioactive source at $t = 0$ was $1600 \ counts \ s^{-1}$,and at $t = 8 \ s$,it was $100 \ counts \ s^{-1}$. The counting rate observed as $counts \ s^{-1}$ at $t = 6 \ s$ will be

State and derive the law of radioactive decay.

The decay constant of a radioactive element is $0.01 \ s^{-1}$. Its half-life period is ....... $s$.

State the law of radioactive decay.

Vedclass Test Series

Exam Paper Generator

Online Exam Module