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The half life period of a radioactive substance is 60 days. The time taken for $$\frac{7}{8}$$th of its original mass to disintegrate will be:
We are given that the half-life of a radioactive substance is $$T_{1/2} = 60$$ days and we need to find the time taken for $$\frac{7}{8}$$th of the original mass to disintegrate.
If $$\frac{7}{8}$$th of the mass has disintegrated, the remaining mass is:
$$\frac{N}{N_0} = 1 - \frac{7}{8} = \frac{1}{8}$$Using the radioactive decay formula:
$$\frac{N}{N_0} = \left(\frac{1}{2}\right)^{t/T_{1/2}}$$Substituting the values:
$$\frac{1}{8} = \left(\frac{1}{2}\right)^{t/60}$$Since $$\frac{1}{8} = \left(\frac{1}{2}\right)^3$$, we get:
$$\left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^{t/60}$$Comparing the exponents:
$$\frac{t}{60} = 3$$ $$t = 180 \text{ days}$$Therefore, the correct answer is Option C: 180 days.
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