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# Working with a radiactive material

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The half life of Cl is 9.5 x 10(exponent 12)s. How many atoms of Cl are present in a sample with an activity of 1.5 x 10(exponent 9) atoms/s

a. 2.1 x 10(exponent 22) atoms
b. 6.3 x 10(exponent 3) atoms
c. 4.9 x 10(exponent negative 23) atoms
d. 1.4 x 10(exponent 22) atoms
e. 7.0 x 10(exponent 23) atoms

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## SOLUTION This solution is FREE courtesy of BrainMass!

Half life t_1/2 = 9.5 x 10^12 s

means, in this time, the radioactive material will have reduced to half its original quantity.

Given activity is A = 1.5 x 10^9 atoms/second

WE WILL NOW FIND THE DECAY CONSTANT LAMDA (L)

L = 0.693/t_1/2

= 0.693/9.5 x 10^12 s = 0.073x10^-12 /Second

Activity A = L N

where N is the number of atoms

Now N = A/L = 1.5 x 10^9/0.073x10^-12

= 20.55 x 10^21

approximately, the answer is 2.1 x 10^22 atoms