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September 1, 2014

September 1, 2014

Posted by **Sandhya** on Sunday, April 20, 2008 at 1:53pm.

I am not able to get the answer.Please help

- physicsRadioactivity -
**DrBob222**, Sunday, April 20, 2008 at 3:01pmDo you have an answer?

- physicsRadioactivity -
**Sandhya**, Sunday, April 20, 2008 at 8:22pmYes. But it is incorrect

R = 1.34 Bq = 1.34 decays / second Mass of sample = 7.02 g

For 14 C T 1/2 = 5700 years (approximate value)

= 5700 x 365 x 24 x 60 x 60 =1.8 x 10 11 seconds

T 1/2 = 0.6931 / λ

λ = 0.6931 / 1.8 x 10 11 = 3.86 x 10 -12 per second

Number of atoms contained in 7.02 g of carbon (12 C atoms)

= (7.02 x 6.023 x 10 23) /14 = 3.02 x 10 23

14 C atoms / 12 C atoms = 1.23 x 10-12

No of 14 C atoms (radioactive) = No of 12 C atoms x 1.23 x 10-12

N0 = 3.02 x 10 23 x 1.23 x 10-12 = 3.7146 x 10 11

R = Nλ

N = R/λ = 1.34 / 3.86 x 10 -12 = 3.4715 x 10 11

t = 3.323 x 5700 x In (3.4715 x 10 11 / 3.7146 x 10 11 ) = 557 years

- physicsRadioactivity -

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