Question

The law of radioactive decay equation is: (Where, T = Half-life period and $\lambda$ = radioactive disintegration constant)

• A. $T = \frac{\log_e 2}{\lambda}$
• B. $T = \frac{\log_{10} 2}{\lambda}$
• C. $T = \frac{\log_e \lambda}{2}$
• D. $T = \frac{\log_{10} \lambda}{2}$

Hint:

Always look for the Natural Log ($\log_e$). Radioactive decay happens in nature, so it uses the natural base '$e$'. The "2" comes from the fact that it's a half (splitting into 2) life.

Answer 1

The Correct Option is A

Radioactive decay is a random process at the atomic level, but for a large number of atoms, it follows a very precise mathematical law known as the Law of Radioactive Disintegration. 1. The Decay Equation: The number of atoms ($N$) remaining after a time ($t$) is given by: $$N = N_0 e^{-\lambda t}$$ Where $N_0$ is the initial number of atoms and $\lambda$ is the decay constant.

2. Defining Half-Life (T): The half-life is the time required for the number of radioactive atoms to decrease to exactly half of its initial value ($N = N_0 / 2$).

3. Mathematical Derivation: Substituting the half-life condition into the decay equation: $$\frac{N_0}{2} = N_0 e^{-\lambda T}$$ $$\frac{1}{2} = e^{-\lambda T}$$ Taking the natural logarithm ($\ln$ or $\log_e$) of both sides: $$\log_e \left(\frac{1}{2}\right) = -\lambda T$$ $$-\log_e 2 = -\lambda T$$ $$T = \frac{\log_e 2}{\lambda}$$

4. Numerical Value: Since $\log_e 2$ (natural log of 2) is approximately 0.693, the relationship is often written as $T = 0.693 / \lambda$. This formula shows that the half-life is inversely proportional to the disintegration constant; a substance that decays rapidly (large $\lambda$) will have a very short half-life.