The decay constant of a radioactive substance | vedclass

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

Question

(D) Given decay constant $\lambda = 0.173 \, (years)^{-1}.$$1$. Half-life $T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{0.173} \approx 4 \, years.$$

Vedclass · Accepted Answer

$A$ and $C$ both.

Vedclass · Answer

(D) Given decay constant $\lambda = 0.173 \, (years)^{-1}.$$1$. Half-life $T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{0.173} \approx 4 \, years.$$2$. For time $t = 1/\lambda = 1/0.173 \, years,$ the remaining amount is $N = N_0 e^{-\lambda t} = N_0 e^{-1} \approx 0.37 N_0.$Thus,the decayed amount is $N_0 - 0.37 N_0 = 0.63 N_0,$ which is $63\%.$ So,option $A$ is correct.$3$. After $8 \, years,$ which is $2 	imes T_{1/2},$ the remaining amount is $N = N_0 (1/2)^2 = N_0/4.$ So,option $C$ is correct.Therefore,both $A$ and $C$ are correct.

The decay constant of a radioactive substance is $0.173 \, (years)^{-1}.$ Therefore:

Similar Questions

State the law of radioactive decay.

Some nuclei of a radioactive material are undergoing radioactive decay. The time gap between the instances when a quarter of the nuclei have decayed and when half of the nuclei have decayed is given as: (where $\lambda$ is the decay constant)

Carbon dating is best suited for determining the age of fossils if their age in years is of the order of

The decay constants of two radioactive elements are $15x$ and $3x$ respectively. Initially,they have the same number of nuclei. After a time of $\frac{1}{6x}$,the ratio of the number of their nuclei will be ........

$C^{14}$ has a half-life of $5700$ years. At the end of $11400$ years,the actual amount left is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module