Question

The half life of 231Pa is 3.25 x 10^4

years. How much of an initial 10.40
microgram sample remains after 3.25 x
10^5 years?​

Answer 1

Answer: 0.0102 micrograms

Step-by-step explanation:

Given t1/2 for 231Pa = 3.25 x 10^4 yr

The radioactive disintegration constant is related by the equation as,

λ = 0.693/t1/2

λ = 0.693/(3.25 x 10^4) yr-1

λ = 2.1323 x 10^-5 yr^-1

Now radioactive disintegration follows first order rate law so we get a equation,

N(t) = N0e^-λt

Where N0 = Amount initially present = 10.40 microgram

Nt = amount present after time t = need to be calculate

λ = radioactive disintegration constant = 2.1323 x 10^-5 yr^-1

t= time = 3.25 x 10^5 yr

Now from the above equation we get,

N(t) = N0e^-λt

Nt = 10.40e^(2.1323^10-5 x 3.25^105)

Nt = 10.40 x e^-6.9299

Nt = 10.40 x 9.78 x 10^-4

Nt = 0.0102 micrograms