Question

The half time of a radioactive substance is \(20 \mathrm{~min}\), difference between Points of time when it is \(33 \%\) disintegrated and \(67 \%\) disintegrated is approximately (A) \(10 \mathrm{~min}\) (B) \(20 \mathrm{~min}\) (C) \(40 \mathrm{~min}\) (D) \(30 \mathrm{~min}\)

Short answer

The time difference between the points of 33% and 67% disintegration is approximately 20 minutes, which corresponds to option (B).

Step-by-step solution

Write down the formula for radioactive decay

The formula for radioactive decay is given by: \(N(t) = N_0 e^{-\lambda t}\) where: - \(N(t)\) is the amount of the substance at time t, - \(N_0\) is the initial amount of the substance, - \(e\) is the base of natural logarithms, - \(\lambda\) is the decay constant, and - \(t\) is the time elapsed.

Calculate decay constant (\(\lambda\))

To find the decay constant, we will use the formula for half-life: \(t_{1/2} = \frac{\ln 2}{\lambda}\) Since half-life is given as 20 minutes, we have: \(\lambda = \frac{\ln 2}{20}\)

Determine the time for 33% and 67% disintegration

Now, we need to find the time when the substance is 33% and 67% disintegrated. For 33% disintegrated, we have \(N(t) = 0.67 N_0\), which means that 67% of the substance is remaining. We plug this into the decay formula: \(0.67 N_0 = N_0 e^{-\lambda t_1}\) Divide both sides by \(N_0\): \(0.67 = e^{-\frac{\ln 2}{20} t_1}\) Take the natural logarithm of both sides: \(\ln 0.67 = -\frac{\ln 2}{20} t_1\) Solving for \(t_1\): \(t_1 = \frac{20 (\ln 0.67)}{-\ln 2}\) For 67% disintegrated, we have \(N(t) = 0.33 N_0\), which means that 33% of the substance remains. We plug this into the decay formula: \(0.33 N_0 = N_0 e^{-\lambda t_2}\) Divide both sides by \(N_0\): \(0.33 = e^{-\frac{\ln 2}{20} t_2}\) Take the natural logarithm of both sides: \(\ln 0.33 = -\frac{\ln 2}{20} t_2\) Solving for \(t_2\): \(t_2 = \frac{20 (\ln 0.33)}{-\ln 2}\)

Calculate the time difference

Now that we have both times, we can find the difference: \(\Delta t = t_2 - t_1 = (\frac{20 (\ln 0.33)}{-\ln2}) - (\frac{20 (\ln 0.67)}{-\ln2})\) Simplify the expression: \(\Delta t = \frac{20(\ln 0.33 - \ln 0.67)}{-\ln 2}\) \(\Delta t = 20 \cdot \frac{\ln(\frac{0.33}{0.67})}{\ln 2}\) Evaluate the expression: \(\Delta t \approx 20.021\) The time difference between the points of 33% and 67% disintegration is approximately 20 minutes, which corresponds to option (B).