Question

The decay constant of radium is $4.28\times10^{-4}$ per year. Its half-life will be:
A. 2000 years.
B. 1240 years.
C. 63 years.
D. 1620 years.

Answer 1

Hint: Here $\lambda$ is already given. We need to use the relationship between half-life and the decay constant, which is our required formula, to get the answer to the value of half-life. The basic concept of this question is the law of radioactive decay.

Formula used:
$T_{1/2}=\dfrac{\ln\lgroup2\rgroup}{\lambda}$
$\ln\lgroup2\rgroup=0.693$

Complete step by step solution:
Considering the information in the question,
$\lambda=4.28\times10^{-4}$ per year.
According to the radioactive decay law, the likelihood that a nucleus would decay per unit of time is a constant that is independent of time, and this constant is symbolized by $\lambda$. A characteristic constant for decaying samples, the half-life is the amount of time required for a sample to decay to its half amount.

Mathematically speaking, the half-life of radioactive materials and the decay constant are connected. The equation, which describes the relationship between the half-life $T_{1/2}$ and the decay constant $\lambda$ is given below:
$T_{1/2}=\dfrac{\ln\lgroup2\rgroup}{\lambda}$
Substituting the above equation with $\ln\lgroup2\rgroup=0.693$ and $\lambda=4.28\times10^{-4}$ we get,
$\Rightarrow T_{1/2}=\dfrac{0.693}{4.28\times~10^{-4}}$
$\therefore T_{1/2}=1620$ years

Hence, the correct option is D.

Note: Some quantities experience simultaneous exponential growth and decline. In this instance, the relationship between the actual half-life $T_{1/2}$ and the half-lives $t_1$ and $t_2$ that the quantity would have if each of the decay processes operated independently is as follows:
$\dfrac{1}{T_{1/2}}=\dfrac{1}{t_{1}}+\dfrac{1}{t_{2}}$
The equivalent formula for three or more processes is:
$\dfrac{1}{T_{1/2}}=\dfrac{1}{t_{1}}+\dfrac{1}{t_{2}}+\dfrac{1}{t_{3}}+\dotsm$