A radioactive material decays by simultaneous | vedclass

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

Question

(A) The rate of decay is given by $\frac{-dN}{dt} = \lambda_1 N + \lambda_2 N = (\lambda_1 + \lambda_2)N$.Integrating this, we get $\ln(\frac{N}{N_0})

Vedclass · Accepted Answer

$1080$

Vedclass · Answer

(A) The rate of decay is given by $\frac{-dN}{dt} = \lambda_1 N + \lambda_2 N = (\lambda_1 + \lambda_2)N$.Integrating this, we get $\ln(\frac{N}{N_0}) = -(\lambda_1 + \lambda_2)t$.Given half-lives $T_1 = 1620 \, years$ and $T_2 = 810 \, years$, the decay constants are $\lambda_1 = \frac{\ln 2}{1620}$ and $\lambda_2 = \frac{\ln 2}{810}$.We want to find $t$ when $\frac{N}{N_0} = \frac{1}{4}$, so $\ln(\frac{1}{4}) = -(\lambda_1 + \lambda_2)t$.$-2 \ln 2 = -(\frac{\ln 2}{1620} + \frac{\ln 2}{810})t$.Dividing by $-\ln 2$, we get $2 = (\frac{1}{1620} + \frac{1}{810})t$.$2 = (\frac{1 + 2}{1620})t = \frac{3}{1620}t = \frac{1}{540}t$.Therefore, $t = 2 	imes 540 = 1080 \, years$.

$A$ radioactive material decays by simultaneous emission of two particles with respective half-lives $1620 \, years$ and $810 \, years$. The time, in years, after which one-fourth of the material remains is

Similar Questions

Which of the following is absent in polluted water?

Which of the following can show optical isomerism?

Carbon cannot be used in the reduction of $Al_2O_3$ because :

$A$ millivoltmeter of $25 \, mV$ range is to be converted into an ammeter of $25 \, A$ range. The value (in $\Omega$) of the necessary shunt will be:

Carbohydrates,the most abundant biomolecules on Earth,are produced by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module