Question

A radioactive isotope has an activity of 8.58×104 Bq initially. After 3.85 hours the activity is 5.15×104 Bq. What is the half-life of the isotope? Tries 0/20 What is the activity after an additional 3.85 hours?

Answer 1

Initial activity of the radioactive isotope = N₀ = 8.58 x 10⁴ Bq

Time period = T₁ = 3.85 hours

Activity of radioactive isotope after 3.85 hours = N₁ = 5.15 x 10⁴ Bq

Decay constant =

By radioactive decay law,

-3.85 = -0.5104

= 0.1326

Half life of the isotope = T

T = 5.227 hours

Another 3.85 hours have passed.

Time period = T₂ = 3.85 + 3.85 = 7.7 hours

Activity of radioactive isotope after 7.7 hours = N₂

N₂ = 3.09 x 10⁴ Bq

Half-life of the isotope = 5.227 hours

Activity after an additional 3.85 hours have passed = 3.09 x 10⁴ Bq