Assuming that there is no decay during the interval,
Nt = 105 × 60 × q₀.
N₀
Thus -------- = 480,000.
q₀
Making the small correction for the decay of activity during the interval,
N₀
---- = 477,000.
q₀
We have previously shown that from the theory
Assuming that there is no decay during the interval,
Nt = 105 × 60 × q₀.
N₀
Thus -------- = 480,000.
q₀
Making the small correction for the decay of activity during the interval,
N₀
---- = 477,000.
q₀
We have previously shown that from the theory