The decay constant of the parent nuclide in the uranium series is $\lambda$ . Then the decay constant for the stable end product of the series will be?
\[
A.{\text{ }}\dfrac{\lambda }{{238}} \\
B.{\text{ }}\dfrac{\lambda }{{206}} \\
C.{\text{ }}\dfrac{\lambda }{{208}} \\
D.{\text{ Zero}} \\
\]
Answer
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Hint: In order to deal with this question first we will define the term decay constant further we will come to the question by using the stable condition of the decay constant and according to this condition we will determine the required answer.
Formula used- $\lambda = \dfrac{1}{T}$
Complete step-by-step answer:
Decay constant: The continuous ratio of a radionuclide's number of atoms decaying over a given period of time relative to the cumulative number of atoms of the same form present at the beginning of that period. The law on nuclear decay states that the probability that a nucleus will decay per unit time is a constant, regardless of time. This constant is called the constant of decay, and is referred to as $\lambda $ "lambda".
We know that the decay constant is given as:
Decay constant $\lambda = \dfrac{1}{T}$ , where T is time taken by the initial amount to fall to $\dfrac{1}{e}$ of its initial amount.
As a stable nuclei will never decay at all so time taken to fall to $\dfrac{1}{e}$ of its initial value will be $\infty $ whose inverse is zero
So decay constant for stable nuclei is:
Decay constant $\lambda = \dfrac{1}{\infty } = 0$
Hence, the decay constant for the stable end product of the series will be zero.
So, the correct answer is option D.
Note- This constant of decline also called constant likelihood will vary considerably between various types of nuclei, contributing to the very different rates of decline observed. In time, the radioactive decay of any number of atoms (mass) is exponential. The decay rate is completely different from temperature. Many studies have shown that alpha and beta declines have not been influenced by environmental influences such as temperature, air pressure or organic content.
Formula used- $\lambda = \dfrac{1}{T}$
Complete step-by-step answer:
Decay constant: The continuous ratio of a radionuclide's number of atoms decaying over a given period of time relative to the cumulative number of atoms of the same form present at the beginning of that period. The law on nuclear decay states that the probability that a nucleus will decay per unit time is a constant, regardless of time. This constant is called the constant of decay, and is referred to as $\lambda $ "lambda".
We know that the decay constant is given as:
Decay constant $\lambda = \dfrac{1}{T}$ , where T is time taken by the initial amount to fall to $\dfrac{1}{e}$ of its initial amount.
As a stable nuclei will never decay at all so time taken to fall to $\dfrac{1}{e}$ of its initial value will be $\infty $ whose inverse is zero
So decay constant for stable nuclei is:
Decay constant $\lambda = \dfrac{1}{\infty } = 0$
Hence, the decay constant for the stable end product of the series will be zero.
So, the correct answer is option D.
Note- This constant of decline also called constant likelihood will vary considerably between various types of nuclei, contributing to the very different rates of decline observed. In time, the radioactive decay of any number of atoms (mass) is exponential. The decay rate is completely different from temperature. Many studies have shown that alpha and beta declines have not been influenced by environmental influences such as temperature, air pressure or organic content.
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