Question

1.2 mg of \[^{239}Pu\] emits \[1.4 \times {10^7}\] particles per minute. What is its half-life?
A.2.372 years
B.3.372 years
C.4.372 years
D.None

Answer 1

Hint: Plutonium, with the symbol Pu and atomic number 94, is a radioactive chemical element. It's a silvery-gray actinide metal that tarnishes in the presence of air and creates a dull covering when oxidised. The element has six allotropes and four oxidation levels in its natural form.

Complete answer: The half-life is the amount of time it takes for a quantity to decline to half its original value. In nuclear physics, the word is used to explain how rapidly unstable atoms disintegrate radioactively and how long stable atoms survive. The phrase is also used to describe any form of exponential or non-exponential decay in general. The biological half-life of medications and other compounds in the human body, for example, is discussed in medical research. Doubling time is the inverse of half-life. The word "half-life" is almost exclusively applied to exponential (radioactive decay, for example) or roughly exponential decay processes (such as biological half-life discussed below). The half-life of a degradation process that is not even close to exponential will alter substantially while it is occurring. In this situation, it is uncommon to speak of half-life in the first place, but people will occasionally refer to the decay as having a "first half-life," "second half-life," and so on, where the first half-life is defined as the time required to decay from the initial value to 50%, the second half-life is defined as the time required to decay from 50% to 25%, and so on.
$-\dfrac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=\lambda \mathrm{N}$
$0.1 \mathrm{mg}^{239} \mathrm{P} \mathrm{u}=0.1 \times 10^{-3} \mathrm{~g}=\dfrac{0.1 \times 10^{-3}}{239} \mathrm{~mol}$
$\therefore \mathrm{N}=\dfrac{0.1 \times 10^{-3} \times 6.02 \times 10^{2 \mathrm{~s}}}{239}$ atoms
$\therefore 1.4 \times 10^{7}=\dfrac{2 \times 0.1 \times 10^{-3} \times 6.02 \times 10^{23}}{239}$
$\therefore \lambda=5.56 \times 10^{-11} \mathrm{~min}^{-1}$
$\therefore \mathrm{t}_{50}=\dfrac{0.698}{i}=7.481 \times 10^{11} \mathrm{~s}$
$=2.372 \mathrm{years}$
Hence option A is correct.

Note:
The original phrase, half-life time, was used in the early 1950s to refer to Ernest Rutherford's discovery of the concept in 1907. By quantifying the decay time of radium to lead-206, Rutherford used the notion of a radioactive element's half-life to studies of age determination of rocks. The half-life of an exponentially decaying quantity is constant across its lifespan, and it is a characteristic unit for the exponential decay equation. The reduction of a quantity as a function of the number of half-lives elapsed is shown in the table below.