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? a) 5.0 min after it formsb) 1.0 h after it forms

2.2 % and 0 %

Explanation:

The equation we will be using to solve this question is:

N/N₀ = e⁻λ t

where N₀ : Number of paricles at t = 0

N = Number of particles after time t

λ = Radioactive decay constant

e = Euler's constant

We are not given λ, but it can be determined from the half life with the equation:

λ = 0.693 / t 1/2 where t 1/2 is the half-life

Substituting our values:

λ = 0.693 / 55 s = 0.0126/s

a) For t = 5 min = 300 s

N / N₀ = e^ - (0.0126/s x 300 s) = e^-3.8 = 0.022 = 2.2 %

b) For t = 1 hr = 3600 s

N / N₀ = e^ - (0.0126/s x 3600 s) = 2.9 x 10 ⁻²⁰ = 0 % (For all practical purposes)