Question

The count rate from a radio active sample falls from $4.0\times {10}^6$ per second to $1.0\times {10}^6$ per second in 20hours. What will be the count rate in 100 hours after the beginning?

• A. $3.9\times {10}^9$per second
• B. $3.9\times {10}^{12}$per second
• C. $3.9\times {10}^3$per second
• D. $3.9\times {10}^6$per second

Answer 1

Answer: (C)

$3.9\times {10}^3$per second

For nuclear reaction,

$K=\frac{2.303}{t}log\frac{A_o}{A}$

$\Rightarrow K=\frac{2.303}{20}log\frac{4\times {10}^6}{1\times {10}^6}$

$\Rightarrow K=0.0693$

After $100$ hours,

$K=\frac{2.303}{t}log\frac{A_o}{A}$

$\Rightarrow 0.0693=\frac{2.303}{100}log\frac{4\times {10}^6}{x}$

$\Rightarrow 3.01=log\frac{4\times {10}^6}{x}$

$x=\frac{4\times {10}^6}{3.01}=3.9\times {10}^{3}persecond$