Question

Pre-calc help please?

A certain radioactive isotope is a by-product of some nuclear reactors. Due to an explosion, a nuclear reactor experiences a massive leak of this radioactive isotope. Fortunately, the isotope has a very short half-life of 10 days. After 2 days, about what percent of the original amount of the isotope released by the explosion remains?

Answer 1

@Directrix @ILovePuppiesLol could you help me please?

Answer 2

The number of atoms \(N\) as a function of time past \(t\),
is equal to the initial number \(N_0\)
times a half to the power of
the time past, divided by the half-life t1/2.
\[\large N(t)=N_0(\tfrac12)^{\frac t{t_{1/2}}}\]
The ratio \(N(t)/N_0\) is the ratio of the remaining isotope after a time \(t\).

So. substitute \(t=2\), and \(t_{1/2}=10\),
into:
\[\large\frac{N(t)}{N_0}=(\tfrac12)^{\frac t{t_{1/2}}}\]

Answer 3

@hockeychick23