Let's evaluate the statements one by one to determine which is not true for radioactive decay.
Statement 1: The amount of radioactive substance remaining after three half-lives is \( \frac{1}{8} \)th of the original amount.
This statement is true. Radioactive substances decay by half over each half-life period. After one half-life, \( \frac{1}{2} \) remains; after two half-lives, \( \frac{1}{4} \) remains; and after three half-lives, \( \frac{1}{8} \) remains. Mathematically, this is expressed as \( \frac{1}{2^n} \) for \( n \) half-lives.
Statement 2: Decay constant does not depend upon temperature.
This statement is true. The decay constant (\( \lambda \)) is a characteristic of each radioactive isotope and does not change with temperature or pressure.
Statement 3: Decay constant increases with increase in temperature.
This statement is false. As established, the decay constant is independent of temperature changes, making this statement incorrect in the context of radioactive decay.
Statement 4: Half-life is \( \frac{\ln(2)}{\text{rate constant}} \).
This is a true statement. The half-life \( t_{\frac{1}{2}} \) is related to the decay constant by the formula \( t_{\frac{1}{2}} = \frac{\ln(2)}{\lambda} \).
Therefore, the incorrect statement regarding radioactive decay is: "Decay constant increases with increase in temperature."
Radioactive decay is a nuclear process and is independent of temperature, pressure, or chemical state of the substance.
The decay constant (λ) and half-life (t1/2) are related as: \[ t_{1/2} = \frac{\ln 2}{\lambda} \]
Hence:
✅ Option 3: Decay constant increases with increase in temperature. (This statement is not true.)
If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then \(a + b + ab\) is equal to:
Given three identical bags each containing 10 balls, whose colours are as follows:
| Bag I | 3 Red | 2 Blue | 5 Green |
| Bag II | 4 Red | 3 Blue | 3 Green |
| Bag III | 5 Red | 1 Blue | 4 Green |
A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from Bag I is $ p $ and if the ball is Green, the probability that it is from Bag III is $ q $, then the value of $ \frac{1}{p} + \frac{1}{q} $ is:
If \( \theta \in \left[ -\frac{7\pi}{6}, \frac{4\pi}{3} \right] \), then the number of solutions of \[ \sqrt{3} \csc^2 \theta - 2(\sqrt{3} - 1)\csc \theta - 4 = 0 \] is equal to ______.