Question

The radioactive isotope (82 Sr) has a half-life of 25.4 days. A sample containing this isotope has an initial activity at (t = 0) of 4.5 x 10^8 Bq. Calculate the number of nuclei that will decay in the time interval between t1 = 34.0 hours and t2 = 50.0 hours.

Answer 1

T = half life period = 25.4 days = 25.4 x 24 h = 609.6 h

decay constant is given as

= 0.693/T = 0.693 /609.6 = 0.00114

A_o = initial activity = 4.5 x 10⁸ Bq

N_o = initial number of nuclei

we know that

A_o = N_o  

4.5 x 10⁸ = (0.00114) N_o  

N_o = 3.94737 x 10¹¹

N₁ = Number of nuclei at t₁ = 34 h ,

Using the equation

N₁ = N_o 2^-t₁/T

N₁ = (3.94737 x 10¹¹ ) 2^-34/609.6

N₁ = 3.798 x 10¹¹

N₂ = Number of nuclei at t₂ = 50 h

Using the equation

N₂ = N_o 2^-t₁/T

N₂ = (3.94737 x 10¹¹ ) 2^-50/609.6

N₂ = 3.7292 x 10¹¹

N = Number of nuclei decayed = N₁ - N₂ = 3.798 x 10¹¹ - 3.7292 x 10¹¹ = 6.88 x 10⁹