The mean lives of a radioactive sample are 30 | vedclass

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

Question

(A) The decay constant for $\alpha$-emission is $\lambda_1 = \frac{1}{	au_1} = \frac{1}{30} \, 	ext{year}^{-1}$.The decay constant for $\beta$-emiss

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(A) The decay constant for $\alpha$-emission is $\lambda_1 = \frac{1}{	au_1} = \frac{1}{30} \, 	ext{year}^{-1}$.The decay constant for $\beta$-emission is $\lambda_2 = \frac{1}{	au_2} = \frac{1}{60} \, 	ext{year}^{-1}$.Since the sample decays by both processes simultaneously,the effective decay constant is $\lambda_{eff} = \lambda_1 + \lambda_2 = \frac{1}{30} + \frac{1}{60} = \frac{2+1}{60} = \frac{3}{60} = \frac{1}{20} \, 	ext{year}^{-1}$.The effective mean life is $	au_{eff} = \frac{1}{\lambda_{eff}} = 20 \, 	ext{years}$.The half-life is $T_{1/2} = 	au_{eff} \ln(2) = 20 	imes 0.693 = 13.86 \, 	ext{years}$.For the sample to reduce to one-fourth of its initial amount,the time $t$ required is given by $N(t) = N_0 (1/2)^n$,where $n$ is the number of half-lives.Since $(1/2)^2 = 1/4$,we need $n = 2$ half-lives.Therefore,$t = 2 	imes T_{1/2} = 2 	imes 13.86 \approx 27.72 \, 	ext{years}$,which is approximately $28 \, 	ext{years}$.

Similar Questions

$A$ radioactive material has an initial amount $16 \, gm$. After $120 \, days$ it reduces to $1 \, gm$. The half-life of the radioactive material is .......... $days$.

If the mass of a radioactive sample is doubled,the activity of the sample and the disintegration constant of the sample are respectively

$A$ source contains two phosphorus radionuclides $_{15}^{32} P \left(T_{1/2} = 14.3 \ d\right)$ and $_{15}^{33} P \left(T_{1/2} = 25.3 \ d\right)$. Initially,$10\%$ of the decays come from $_{15}^{33} P$. How long must one wait until $90\%$ of the decays come from $_{15}^{33} P$?

The half-life of a radioactive nuclide is

Vedclass Test Series

Exam Paper Generator

Online Exam Module