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 of bromine-82, \( t_{1/2} = 36 \, \text{hours} \).  
\(t_{1/2} = \frac{0.693}{K}\)
\(K = \frac{0.693}{36} = 0.01925 \, \text{hr}^{-1}\)

For a 1\(^\text{st}\)-order reaction, the kinetic equation is:  
\(t = \frac{2.303}{K} \log \frac{a}{a-x}\)

For \( t = 1 \, \text{day} \, (t = 24 \, \text{hr}) \):  
\(\log \frac{a}{a-x} = \frac{t \times K}{2.303}\)
\(\log \frac{a}{a-x} = \frac{24 \, \text{hr} \times 0.01925 \, \text{hr}^{-1}}{2.303}\)
\(\log \frac{a}{a-x} = 0.2006\)

Now,  
\(\frac{a}{a-x} = \text{antilog} \, (0.2006)\)

\(\frac{a}{a-x} = 1.587\)

If \( a = 1 \):  

\(\frac{1}{1-x} = 1.587 \implies 1-x = 0.6301\)

Thus, the fraction remaining after one day is:  

\(1-x = 0.6301 \approx 63\%\)

The Correct Answer is: 63%