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April 21, 2014

April 21, 2014

Posted by **Sandhya** on Saturday, April 19, 2008 at 2:09pm.

2) A specimen taken from the wrappings of a mummy contains 7.02 g of carbon and has an activity of 1.34 Bq. How old is the mummy? Determine its age in years assuming that in living trees the ratio of 14C/12C atoms is 1.23E-12.

Please check my answers

1)

For 14 C T 1/2 = 5570 years N/N0 = 11.9/100 = 0.119

N / N0 = e- λ t 2.303 x In 0.119 = -λt λ =0.6931 / 5570 =1.244 x 10 -4

t = (2.303 x In0.119 x 5570) / 0.6931 = 17110 years

2)

R = 1.34 Bq = 1.34 decays / second

14 C atoms / 12 C atoms = 1.23 x 10-12 = N0/N N/N0 = 10 12 /1.2

Number of atoms contained in 7.02 g of carbon

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

R = Nλ λ = 1.34 / 3.02 x 10 23 = 0.4437 x 10 -23

2.303 x In (10 12 /1.23 ) = -0. 4437 x 10 -23 x t

t = 61.82 x 10 23 seconds = 1.96 x 10 17 years

- Physics -
**Sushma**, Wednesday, December 17, 2008 at 12:41pmT 1/2 of C14=5730 yrs.

Here is the answer for the 1st one, hope it will be helpfull.

Lamda= ln2/(T1/2)

= 0.693/5730=1.209E-4

N/N0= 11.9/100 = 0.119

N=N0 e^-lamda*t

N/N0=e^lamda*t

ln(0.119)= 1.209E-4t

t=ln(0.119)/(1.209E-4)

t=17606.5 yrs.

- Physics -
**Sushma Rai**, Wednesday, December 17, 2008 at 12:46pmT 1/2 of C14=5730 yrs.

Here is the answer for the 1st one, hope it will be helpfull.

Lamda= ln2/(T1/2)

= 0.693/5730=1.209E-4

N/N0= 11.9/100 = 0.119

N=N0 e^-lamda*t

N/N0=e^lamda*t

ln(0.119)= 1.209E-4t

t=ln(0.119)/(1.209E-4)

t=17606.5 yrs.

(ps: this is also for my Montgomery College, Takoma Park college Dr. Shaleen Shukla PH 204 class)

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