**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%