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. To find the fraction remaining (F) after 24 hours (1 day), we use the decay formula:

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

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

Now substitute into the formula:
\[ F = \left( \frac{1}{2} \right)^{\frac{2}{3}}. \]

Using logarithmic properties, we convert to logarithmic form:
\[ \log_{10}F = \left( \frac{2}{3} \right) \log_{10}\left( \frac{1}{2} \right). \]

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

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

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

To express this fraction as requested, multiply by \(10^2\):
\[ 0.6301 \times 10^2 \approx 63.01. \]

Answer 2

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%