99% of the nuclei of a radioactive element de | vedclass

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

Question

(A) Let $N_0$ be the initial number of nuclei and $N$ be the number of undecayed nuclei after time $t$.Given that $99\%$ of the nuclei have decayed,th

Vedclass · Accepted Answer

$6	au_{1/2}$ and $7	au_{1/2}$

Vedclass · Answer

(A) Let $N_0$ be the initial number of nuclei and $N$ be the number of undecayed nuclei after time $t$.Given that $99\%$ of the nuclei have decayed,the number of undecayed nuclei is $N = N_0 - 0.99N_0 = 0.01N_0$.Using the radioactive decay law: $N = N_0(1/2)^n$,where $n = t/	au_{1/2}$ is the number of half-lives.Substituting the values: $0.01N_0 = N_0(1/2)^n \Rightarrow (1/2)^n = 0.01 \Rightarrow 2^n = 100$.Taking the logarithm on both sides: $n \log_{10}(2) = \log_{10}(100)$.$n 	imes 0.3010 = 2$.$n = 2 / 0.3010 \approx 6.64$.Since $n = t/	au_{1/2}$,we have $t = n 	imes 	au_{1/2} \approx 6.64 	au_{1/2}$.This value lies between $6	au_{1/2}$ and $7	au_{1/2}$.

$99\%$ of the nuclei of a radioactive element decay between which of the following times?

Similar Questions

The activity of a radioactive sample reduces from $A_0$ to $\frac{A_0}{\sqrt{3}}$ in $1$ hour. What will be the activity after $3$ hours more?

$A$ radioactive substance emits two particles with half-lives of $1620$ years and $810$ years respectively. After how much time will one-fourth of the substance remain?

Give the equation form of the exponential law of radioactive decay.

The decay constants of a radioactive substance for $\alpha$ and $\beta$ emission are $\lambda_{\alpha}$ and $\lambda_{\beta}$ respectively. If the substance emits $\alpha$ and $\beta$ simultaneously,then the average half-life of the material will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module