The number of atoms of a radioactive substanc | 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 $N(t) = N_{0} e^{-\lambda t}$, where $\lambda = \frac{\ln 2}{T}$.Let $t_{1}$ be the time when the number of

Vedclass · Accepted Answer

$T \frac{\log 5}{\log 2}$

Vedclass · Answer

(A) The radioactive decay law is given by $N(t) = N_{0} e^{-\lambda t}$, where $\lambda = \frac{\ln 2}{T}$.Let $t_{1}$ be the time when the number of atoms is $N_{0}/2$ and $t_{2}$ be the time when the number of atoms is $N_{0}/10$.For $N(t_{1}) = N_{0}/2$, we have $\frac{N_{0}}{2} = N_{0} e^{-\lambda t_{1}} \Rightarrow e^{\lambda t_{1}} = 2 \Rightarrow \lambda t_{1} = \ln 2$.For $N(t_{2}) = N_{0}/10$, we have $\frac{N_{0}}{10} = N_{0} e^{-\lambda t_{2}} \Rightarrow e^{\lambda t_{2}} = 10 \Rightarrow \lambda t_{2} = \ln 10$.The time interval required is $\Delta t = t_{2} - t_{1} = \frac{\ln 10 - \ln 2}{\lambda} = \frac{\ln(10/2)}{\lambda} = \frac{\ln 5}{\lambda}$.Since $\lambda = \frac{\ln 2}{T}$, we substitute $\lambda$ to get $\Delta t = \frac{\ln 5}{(\ln 2 / T)} = T \frac{\ln 5}{\ln 2} = T \log_{2} 5$.

The number of atoms of a radioactive substance of half-life $T$ is $N_{0}$ at $t=0$. The time necessary to decay from $N_{0} / 2$ atoms to $N_{0} / 10$ atoms will be

Similar Questions

The half-life of radium is $1620$ years and its atomic weight is $226 \ kg/kmol$. The number of atoms that will decay from its $1 \ g$ sample per second will be (Avogadro's number $N_A = 6.02 \times 10^{26} \ atoms/kmol$)

$A$ freshly prepared radioactive source of half-life $2$ hours $30$ minutes emits radiation which is $64$ times the permissible safe level. The minimum time,after which it would be possible to work safely with the source,will be in hours.

$A$ radioactive sample is an $\alpha$-emitter with a half-life of $138.6$ days. $A$ student observes its activity to be $2000$ disintegrations per second. The number of radioactive nuclei for this given activity is:

The half-life of $Bi^{210}$ is $5$ days. If we start with $50,000$ atoms of this isotope,the number of atoms left over after $10$ days is:

At any instant,two elements $X_1$ and $X_2$ have the same number of radioactive atoms. If the decay constants of $X_1$ and $X_2$ are $10\lambda$ and $\lambda$ respectively,then the time when the ratio of their atoms becomes $\frac{1}{e}$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module