Let N_{beta} be the number of beta particles | vedclass

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

Question

(B) The initial number of nuclei $N_0$ is given by $N_0 = \frac{m}{M} 	imes N_A = \frac{1}{24} 	imes 6.023 	imes 10^{23} \approx 2.51 	imes 10^{22

Vedclass · Accepted Answer

$7.5 	imes 10^{21}$

Vedclass · Answer

(B) The initial number of nuclei $N_0$ is given by $N_0 = \frac{m}{M} 	imes N_A = \frac{1}{24} 	imes 6.023 	imes 10^{23} \approx 2.51 	imes 10^{22}$.The number of nuclei decayed in time $t$ is $N_{\beta} = N_0(1 - e^{-\lambda t})$.Given $t = 7.5 \, hrs$ and $T_{1/2} = 15 \, hrs$,the decay constant $\lambda = \frac{\ln 2}{T_{1/2}}$.Thus,$N_{\beta} = N_0(1 - e^{-\frac{\ln 2}{15} 	imes 7.5}) = N_0(1 - e^{-\frac{\ln 2}{2}}) = N_0(1 - 2^{-1/2})$.Using $2^{-1/2} \approx 0.707$,we get $N_{\beta} = 2.51 	imes 10^{22} 	imes (1 - 0.707) = 2.51 	imes 10^{22} 	imes 0.293 \approx 7.35 	imes 10^{21}$.Rounding to the nearest option,$N_{\beta} \approx 7.5 	imes 10^{21}$.

Let $N_{\beta}$ be the number of $\beta$ particles emitted by $1 \, g$ of $^{24}Na$ radioactive nuclei (half-life $= 15 \, hrs$) in $7.5 \, hrs$. $N_{\beta}$ is close to (Avogadro number $= 6.023 \times 10^{23} \, mol^{-1}$)

Similar Questions

$A$ radioactive sample $S_{1}$ having the activity $A_{1}$ has twice the number of nuclei as another sample $S_{2}$ of activity $A_{2}$. If $A_{2} = 2 A_{1}$,then the ratio of half-life of $S_{1}$ to the half-life of $S_{2}$ is

The activity of a sample of radioactive material is $A_1$ at time $t_1$ and $A_2$ at time $t_2$ $(t_2 > t_1)$. If its mean life is $T$,then which of the following is true?

At time $t=0$,a container has $N_{0}$ radioactive atoms with a decay constant $\lambda$. In addition,$c$ number of atoms of the same type are being added to the container per unit time. How many atoms of this type are there at $t=T$?

An ancient discovery found a sample where $75 \%$ of the original carbon $(C^{14})$ remains. The age of the sample is: $\left(T_{1/2}(C^{14}) = 5730 \text{ years}, \ln 0.5 = -0.7, \ln 0.75 = -0.3\right)$ (in $\text{ years}$)

$A$ radioactive substance has an average life of $5$ hours. In a time of $5$ hours,

Vedclass Test Series

Exam Paper Generator

Online Exam Module