Question

The count rate of radioactive sample falls from $8.0\times {10}^6$ per second to $1.0\times {10}^6$ per second in 12 hours. What will be the count rate after 20 hours from the beginning?

• A. $25\times {10}^4$
• B. $4\times {10}^4$
• C. $25\times {10}^6$
• D. $3\times {10}^4$

Answer 1

The correct option is A $25\times {10}^4$

$A_0=8\times {10}^{6}A={10}^{6}A=A^0\{\frac{1}{2}{\}}^n{10}^6=8\times {10}^6(\frac{1}{2}{)}^n{2}^n=8n=3$
12 hours = 3 half lives
$12=3\times t_{\frac{1}{2}}{t}_{\frac{1}{2}}=4hour$
20 hours = 5 half lives
$A=A_0\{\frac{1}{2}{\}}^n=\frac{8\times {10}^6}{2^5}=25\times {10}^4$