Question

The half-life of a radioactive substance is 30 min. The time for disintegration \(\frac{7}{8}\)^th part of its original mass will be:

• A. T
• B. 2T
• C. 3T
• D. 8T

Answer 1

The Correct Option is C

To solve the problem of determining the time required for the disintegration of \(\frac{7}{8}\)^th part of a radioactive substance, we will use the concept of radioactive decay and half-life.

The half-life of a radioactive substance is the time required for its quantity to reduce to half of its initial value. Given:

• The half-life (\(T_{1/2}\)) of the substance is 30 minutes.

We need to find the time taken for the substance to decay to \(\frac{1}{8}\) of its original mass, as disintegration of \(\frac{7}{8}\) implies only \(\frac{1}{8}\) is left.

We can use the formula related to radioactive decay:

\[ N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]

where:

• \(N(t)\) is the remaining quantity of the substance at time \(t\).
• \(N_0\) is the initial quantity.
• \(T_{1/2}\) is the half-life of the substance.

We need to solve for \(t\) when \(N(t) = \frac{1}{8} N_0\).

\[ \frac{1}{8}N_0 = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]

Dividing both sides by \(N_0\), we get:

\[ \frac{1}{8} = \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]

We recognize \(\frac{1}{8}\) is \(\left(\frac{1}{2}\right)^3\), so:

\[ \left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]

Equating the exponents, we obtain:

\[ 3 = \frac{t}{T_{1/2}} \]

Solving for \(t\), we get:

\[ t = 3 \cdot T_{1/2} \]

Substituting \(T_{1/2} = 30 \text{ min}\):

\[ t = 3 \times 30 = 90 \text{ min} \]

Therefore, the time required for the disintegration of \(\frac{7}{8}\)^th part of the original mass is 90 minutes or 3T, where \(T = 30 \text{ min}\).