Half-lives for alpha and beta emission of a r | vedclass

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

Question

(B) When a radioactive material decays through two different processes simultaneously,the effective decay constant $\lambda_{eff}$ is the sum of the i

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(B) When a radioactive material decays through two different processes simultaneously,the effective decay constant $\lambda_{eff}$ is the sum of the individual decay constants: $\lambda_{eff} = \lambda_{\alpha} + \lambda_{\beta}$.Given $T_{\alpha} = 16$ years and $T_{\beta} = 48$ years.The decay constants are $\lambda_{\alpha} = \frac{\ln 2}{16}$ and $\lambda_{\beta} = \frac{\ln 2}{48}$.Thus,$\lambda_{eff} = \frac{\ln 2}{16} + \frac{\ln 2}{48} = \ln 2 \left( \frac{3+1}{48} \right) = \frac{\ln 2}{12}$.The effective half-life $T_{eff} = \frac{\ln 2}{\lambda_{eff}} = 12$ years.We want to find the time $t$ when $3/4$th of the material has decayed,which means $1/4$th of the material remains.Using the radioactive decay law: $\frac{N}{N_0} = \left( \frac{1}{2} \right)^{t/T_{eff}}$.$\frac{1}{4} = \left( \frac{1}{2} \right)^{t/12} \implies \left( \frac{1}{2} \right)^2 = \left( \frac{1}{2} \right)^{t/12}$.Equating the exponents: $2 = \frac{t}{12} \implies t = 24$ years.

Half-lives for $\alpha$ and $\beta$ emission of a radioactive material are $16$ years and $48$ years respectively. When the material decays by giving $\alpha$ and $\beta$ emission simultaneously,the time in which $3/4$th of the material decays is ....... years.

Similar Questions

Define the $SI$ unit of radioactivity.

$A$ radioactive sample at any instant has its disintegration rate $5000$ disintegrations per minute. After $5\, minutes$,the rate becomes $1250$ disintegrations per minute. Then,its decay constant (per minute) is

The fraction $f$ of radioactive material that has decayed in time $t$,varies with time $t$. The correct variation is given by the curve

$A$ radioisotope $X$ with a half-life of $1.4 \times 10^{9} \text{ years}$ decays into $Y$,which is stable. $A$ sample of rock from a cave was found to contain $X$ and $Y$ in the ratio $1:7$. The age of the rock is ........ $\times 10^{9} \text{ years}$.

At a given instant,there are $25\%$ undecayed radioactive nuclei in a sample. After $10 \, s$,the number of undecayed nuclei reduces to $6.25\%$. The mean life of the nuclei is...........$ s$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module