To determine how many atoms of uranium-238 the meteorite initially had, we can use the concept of radioactive decay. Uranium-238 decays to lead-206, and since we know the half-life of uranium-238 is 4.5 billion years, the meteorite has gone through one half-life to reach its current state.
- After one half-life (4.5 billion years), half of the original uranium-238 would have decayed. If the initial number of uranium-238 atoms is \(N_0\), after one half-life (4.5 billion years), the number of uranium-238 atoms remaining would be \(N_0 / 2\), and the number of lead-206 atoms produced would be equal to the number of uranium-238 atoms that have decayed.
Thus, if there are now 78 atoms of lead-206, that means 78 atoms of uranium-238 have decayed.
- Therefore, the number of remaining uranium-238 atoms is:
\[ N_0 / 2 = N_0 - 78 \]
- Rearranging this gives us:
\[ N_0 = 2 \times 78 = 156 \text{ atoms} \]
So the initial number of uranium-238 atoms in the meteorite was 156.
The answer is 156.