Question 139: Atoms & Nuclei

Question

The half life of a radioactive substance is T. The time taken, for disintegrating $$\frac{7}{8}$$th part of its original mass will be:

Accepted Answer

Let's use the formula for the remaining mass of a radioactive substance after a certain time:
$$N(t) = N_0(1/2)^{t/T}$$
where N(t) is the mass at time t, NÃ¢ÂÂ is the initial mass, T is the half-life, and t is the time elapsed.
We are given that $$\frac{7}{8}$$th of the original mass has disintegrated. Therefore, the remaining mass is $$\frac{1}{8}$$th of the original mass:
$$\frac{N(t)}{N_0} = \frac{1}{8}$$
Using the formula, we have:
$$\frac{1}{8} = (1/2)^{t/T}$$
Taking the logarithm of both sides:
$$\log_{1/2}\frac{1}{8} = \frac{t}{T}$$
$$3 = \frac{t}{T}$$
Now, solving for t:
$$t = 3T$$
So, the time taken for disintegrating $$\frac{7}{8}$$th part of the original mass is 3T.

Answer

8T

Answer

3T

Answer

T

Answer

2T

Atoms & Nuclei

The half life of a radioactive substance is T. The time taken, for disintegrating $\frac{7}{8}$th part of its original mass will be:

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