"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:"

Question

Accepted Answer

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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, N₀ 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.

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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: