Question

The half life of a radioactive substance is 5 years. After x years, a given sample of the radioactive substance gets reduced to 6.25% of its initial value. The value of x is ______.

Answer 1

Correct Answer: 20

The half-life of a substance refers to the time required for half of the substance to decay. Given the half-life \(T_{1/2}\) is 5 years, we can use the formula for radioactive decay:

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

where \(N(t)\) is the remaining quantity at time \(t\), and \(N_0\) is the initial quantity.

If the sample is reduced to 6.25% of its initial value, we have:

\(\frac{N(t)}{N_0} = 6.25\% = 0.0625\)

Substitute this into the decay formula:

\(0.0625 = \left(\frac{1}{2}\right)^{\frac{t}{5}}\)

Taking the logarithm of both sides:

\(\log_{10}(0.0625) = \frac{t}{5} \cdot \log_{10}\left(\frac{1}{2}\right)\)

Solving for \(t\):

\(t = 5 \cdot \frac{\log_{10}(0.0625)}{\log_{10}\left(\frac{1}{2}\right)}\)

Now compute the values:

\(\log_{10}(0.0625) \approx -1.2041\) and \(\log_{10}\left(\frac{1}{2}\right) \approx -0.3010\)

\(t = 5 \cdot \frac{-1.2041}{-0.3010} = 5 \cdot 4\)

\(t = 20\)

Thus, the value of \(x\) is 20 years. This falls within the expected range of 20,20.