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Question: The count rate of radioactive sample falls from \(4.0 \times 10^{6}s^{- 1}\) to \(1.0 \times 10^{6}s...

The count rate of radioactive sample falls from 4.0×106s−14.0 \times 10^{6}s^{- 1} to 1.0×106s−11.0 \times 10^{6}s^{- 1} in 20 hours. What will be the count rate after 100 hours from beginning?

A

3.91×103s−13.91 \times 10^{3}s^{- 1}

B

3.91×102s−13.91 \times 10^{2}s^{- 1}

C

3.91×104s−13.91 \times 10^{4}s^{- 1}

D

3.91×106s−13.91 \times 10^{6}s^{- 1}

Answer

3.91×103s−13.91 \times 10^{3}s^{- 1}

Explanation

Solution

: here, R0=4.0×106s−1,R=1.0×106s−1R_{0} = 4.0 \times 10^{6}s^{- 1},R = 1.0 \times 10^{6}s^{- 1}

t=20hours,t′=100hourst = 20hours,t' = 100hours

As, RR0=(12)nor1.0×1064.0×106=(12)n\frac{R}{R_{0}} = \left( \frac{1}{2} \right)^{n}or\frac{1.0 \times 10^{6}}{4.0 \times 10^{6}} = \left( \frac{1}{2} \right)^{n}

Or (12)n=(12)2∴n=2\left( \frac{1}{2} \right)^{n} = \left( \frac{1}{2} \right)^{2}\therefore n = 2

T1/2=tn=202=10hoursT_{1/2} = \frac{t}{n} = \frac{20}{2} = 10hours

Now , n′=t′T1/2=10010=10n' = \frac{t'}{T_{1/2}} = \frac{100}{10} = 10

∴R′=R0(12)n′=4.0×106(12)10=3.91×103s−1\therefore R' = R_{0}\left( \frac{1}{2} \right)^{n'} = 4.0 \times 10^{6}\left( \frac{1}{2} \right)^{10} = 3.91 \times 10^{3}s^{- 1}