Question

The half-life of radio isotopic bromine - 82 is 36 hours. The fraction which remains after one day is ___________ ×10^–2.
(Given antilog 0.2006 = 1.587)

Answer 1

Correct Answer: 63

The half-life (T_1/2) of bromine-82 is 36 hours. The fraction remaining (F) after 24 hours (1 day) is calculated using the decay formula:

\(F = \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}\), with \(t = 24\) hours.

First, determine the ratio \( \frac{t}{T_{1/2}} \):
\[ \frac{t}{T_{1/2}} = \frac{24}{36} = \frac{2}{3}. \]

Substitute this ratio into the decay formula:
\[ F = \left( \frac{1}{2} \right)^{\frac{2}{3}}. \]

To solve for F, convert to logarithmic form using base-10 logarithms:
\[ \log_{10}F = \left( \frac{2}{3} \right) \log_{10}\left( \frac{1}{2} \right). \]

Calculate the logarithm of 1/2:
\[ \log_{10}\left( \frac{1}{2} \right) = -\log_{10}2 \approx -0.3010. \]

Then, calculate \( \log_{10}F \):
\[ \log_{10}F = \left( \frac{2}{3} \right)(-0.3010) \approx -0.2006. \]

Using the antilogarithm:
\[ \text{antilog}(-0.2006) = \frac{1}{1.587} \approx 0.6301. \]

To express this fraction as a percentage, multiply by 100:
\[ 0.6301 \times 10^2 \approx 63.01. \]