A 1.00 * 10^-6 -g sample of nobelium, 254/102 No, has a half-life of 55 seconds after it is formed

sample-nobelium

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A 1.00 * 10^-6 -g sample of nobelium, 254/102 No, has a half-life of 55 seconds after it is formed. What is the percentage of 254/102 No remaining at the following times?
5.0 min after it forms

Answer:

Radioactive decay follows first order kinetics. So,
N/N0 = e^(-k*t)
where N is the amount of radioactive material left
N0 is the initial amount of the material
k is the constant
t is the time given

Also half-life of the sample, T(1/2) = 0.693/k
So, k = 0.693/55 = 0.0126
Now percentage of sample remaining after 5 min, i.e 300 seconds =
N/N0 * 100 = e^(-0.0126 * 300) * 100 = 2.282%

So answer is 2.282%