A radioactive sample is emitting 64 times mor | vedclass

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

Question

(B) The radioactive sample emits $64$ times the non-hazardous limit,so the ratio of current activity to the limit is $N/N_0 = 1/64$.Using the radioact

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) The radioactive sample emits $64$ times the non-hazardous limit,so the ratio of current activity to the limit is $N/N_0 = 1/64$.Using the radioactive decay formula: $N = N_0(1/2)^n$,where $n$ is the number of half-lives.$1/64 = (1/2)^n$ $\Rightarrow (1/2)^6 = (1/2)^n$ $\Rightarrow n = 6$.Given the half-life $t_{1/2} = 2 \ hr$,the total time $T = n 	imes t_{1/2} = 6 	imes 2 = 12 \ hr$.Thus,the sample becomes non-hazardous after $12 \ hr$.

$A$ radioactive sample is emitting $64$ times more radiation than the non-hazardous limit. If its half-life is $2 \ hr$,after what time will it become non-hazardous? $(hr)$

Similar Questions

The half-life period of a radioactive metal is $20 \ days$. What fraction of the metal remains after $80 \ days$?

The half-life of $^{60}Co$ is $7 \ years$. If $1 \ g$ of it decays,the amount of the substance remaining after $28 \ years$ is (in $g$)

The disintegration constant of radium with a half-life of $1600 \ years$ is:

If the amount of radioactive substance is increased three times,the number of atoms disintegrated per unit time would

Decay of $_{92}U^{235}$ is $...$ order reaction.

Vedclass Test Series

Exam Paper Generator

Online Exam Module