A radioactive material has a DK constant of 0.0122. 8, million nuclei ever meaning after 15 seconds of decay what is the number of original nuclei?
To find the original number of nuclei, we need to first understand the radioactive decay process as described by the decay constant (\( \lambda \)) and the decay law.
The decay law for radioactive decay is given by:
\[ N(t) = N_0 e^{-\lambda t} \]
where:
- \( N(t) \) is the number of nuclei remaining after time \( t \),
- \( N_0 \) is the original number of nuclei at \( t = 0 \),
- \( \lambda \) is the decay constant,
- \( e \) is the base of the natural logarithms (approximately equal to 2.71828).
Given:
- The decay constant \( \lambda = 0.0122 \) per second,
- The remaining number of nuclei after 15 seconds \( N(15) = 8,000,000 \),
- \( t = 15 \) seconds.
We can rearrange the decay law to solve for \( N_0 \):
\[ N_0 = \frac{N(t)}{e^{-\lambda t}} \]
Plugging in the values:
\[ N_0 = \frac{8,000,000}{e^{-0.0122 \times 15}} \]
First, we calculate the exponent:
\[ -\lambda t = -0.0122 \times 15 = -0.183 \]
Then, raise \( e \) to the power of \( -0.183 \):
\[ e^{-0.183} \approx 0.8326 \]
Now we can find \( N_0 \):
\[ N_0 = \frac{8,000,000}{0.8326} \]
\[ N_0 \approx 9,610,835 \]
Therefore, the original number of nuclei was approximately 9,610,835.