A radioactive nucleus n_2 has 3 times the dec | vedclass

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

Question

(A) Let the decay constant of $n_1$ be $\lambda_1 = \lambda$. Then the decay constant of $n_2$ is $\lambda_2 = 3\lambda$.The number of nuclei remainin

Vedclass · Accepted Answer

$1/4$

Vedclass · Answer

(A) Let the decay constant of $n_1$ be $\lambda_1 = \lambda$. Then the decay constant of $n_2$ is $\lambda_2 = 3\lambda$.The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.For $n_1$: $N_1(t) = N_0 e^{-\lambda t}$.For $n_2$: $N_2(t) = N_0 e^{-3\lambda t}$.The time $t$ is one half-life of $n_1$,so $t = T_{1/2} = \frac{\ln 2}{\lambda}$.Substituting $t$ into the expressions:$N_1(t) = N_0 e^{-\lambda (\frac{\ln 2}{\lambda})} = N_0 e^{-\ln 2} = N_0 / 2$.$N_2(t) = N_0 e^{-3\lambda (\frac{\ln 2}{\lambda})} = N_0 e^{-3 \ln 2} = N_0 (e^{\ln 2})^{-3} = N_0 (2)^{-3} = N_0 / 8$.The ratio $\frac{N_2(t)}{N_1(t)} = \frac{N_0 / 8}{N_0 / 2} = \frac{2}{8} = \frac{1}{4}$.

$A$ radioactive nucleus $n_2$ has $3$ times the decay constant as compared to the decay constant of another radioactive nucleus $n_1$. If initial number of both nuclei are the same,what is the ratio of number of nuclei of $n_2$ to the number of nuclei of $n_1$,after one half-life of $n_1$?

Similar Questions

The activity of a sample of radioactive material is $A_1$ at time $t_1$ and $A_2$ at time $t_2$ $(t_2 > t_1)$. Its mean life is $T$.

If a radioactive substance reduces to $\frac{1}{16}$ of its original mass in $40$ days,what is its half-life in days?

$A$ neutron beam has a kinetic energy of $0.0837 \, eV$. Its half-life is $693 \, s$ and its mass is $1.675 \times 10^{-27} \, kg$. What fraction of the neutrons will remain undecayed after traveling a distance of $40 \, m$?

$A$ radioactive nucleus decays by two different processes. The half-life of the first process is $5$ minutes and that of the second process is $30\,s$. The effective half-life of the nucleus is calculated to be $\frac{\alpha}{11}\,s$. The value of $\alpha$ is $..............$

The half-life of a radioactive substance is $20 \, \text{minutes}$. The approximate time interval $(t_2 - t_1)$ between the time $t_2$ when $\frac{3}{4}$ of it has decayed and time $t_1$ when $\frac{1}{4}$ of it has decayed is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module