Question

The half life of 9030sr is 28 years. What is the disintegration rate of 15mg of their isotope ?

Answer 1

Answer:

2.13 Ci

Step-by-step explanation:

Half life of ⁹⁰₃₀Sr = 28 years.

Using the formula, λ = 0.693/T

                                   = 0.693/28 *365 * 24 * 60 * 60

                                   = 7.85 * 10⁻¹⁰ S⁻¹.

90 grams of Sr contains 6.023 * 10²³ atoms of that element.

15 mg of Sr contains, N₀ =  6.023 * 10²³ * 15 * 10⁻³ /90

                                        = 1.0038 * 10²⁰ atoms.

Disintegration Rate dN/dt = - λ N₀

                                           = - 7.85 * 10⁻¹⁰ * 1.0038 * 10²⁰

                                           =   7.88 * 10¹⁰ dps or Bq

                                           = 7.88 * 10¹⁰/3.7 * 10¹⁰  Ci

                                            = 2.13 Ci