In a radioactive substance at t = 0,the numbe | vedclass

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

Question

(A) The radioactive decay law is given by $N = N_0 \left( \frac{1}{2} \right)^{t/T}$,where $N_0$ is the initial number of atoms,$N$ is the remaining n

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) The radioactive decay law is given by $N = N_0 \left( \frac{1}{2} ight)^{t/T}$,where $N_0$ is the initial number of atoms,$N$ is the remaining number of atoms,$t$ is the time elapsed,and $T$ is the half-life period.Given: $N_0 = 8 	imes 10^4$,$N = 1 	imes 10^4$,and $T = 3 \ years$.Substituting the values into the formula:$1 	imes 10^4 = 8 	imes 10^4 \left( \frac{1}{2} ight)^{t/3}$$\frac{1}{8} = \left( \frac{1}{2} ight)^{t/3}$Since $\frac{1}{8} = \left( \frac{1}{2} ight)^3$,we have:$\left( \frac{1}{2} ight)^3 = \left( \frac{1}{2} ight)^{t/3}$Comparing the exponents:$3 = \frac{t}{3}$$t = 9 \ years$.

In a radioactive substance at $t = 0$,the number of atoms is $8 \times 10^4$. Its half-life period is $3 \ years$. The number of atoms $1 \times 10^4$ will remain after an interval of ........... $years$.

Similar Questions

What is the mass of $1 \, \text{curie}$ of $U^{234}$?

Consider two nuclei of the same radioactive nuclide. One of the nuclei was created in a supernova explosion $5$ billion years ago. The other was created in a nuclear reactor $5$ minutes ago. The probability of decay during the next time interval is

The activity of a sample of a radioactive material is ${A_1}$ at time ${t_1}$ and ${A_2}$ at time ${t_2}$ $({t_2} > {t_1})$. If its mean life is $T$,then:

$A$ radioactive sample is undergoing $\alpha$ decay. At any time $t_{1}$,its activity is $A$ and at another time $t_{2}$,the activity is $\frac{A}{5}$. What is the average life time for the sample?

Which is the correct expression for half-life?

Vedclass Test Series

Exam Paper Generator

Online Exam Module