Pd-100 has a half life of 3.6 days. If one had 6.02 x 10 to the 23 atoms at the start, how many atoms at the start, how many atoms would be present after 20 days?

Answer:
1.28 × 10²² atoms
Step-by-step explanation:
A common formula for determining the amount of sample remaining (N) in terms of its half-life (t_½) is
N =N₀(½)ⁿ
where n is the number of half-lives
n = t/t_½
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Data:
N₀ = 6.02 × 10²³ atoms
t = 20 da
t_½ = 3.6 da
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The calculation:
N₀ =6.02 × 10²³ atoms
n = 20/3.6
n= 5.56
N = 6.02 × 10²³ × (½)^5.56
N = 6.02 × 10²³ × 0.0212
N = 1.28 × 10²² atoms
1.28 × 10²² atoms of Pd-100 would be present after 20 days.
To calculate the amount of atoms present after a half-life decomposition process, it is necessary to use the following formula:
[tex]N =N_{0} (\frac{1}{2})^{t/t_{1/2}[/tex]
Since the initial amount of mass is equal to [tex]6.02\times10^{23}[/tex] atoms, the half-life is 3.6 days and the total time is 20 days, we just have to put these values in the expression above:
[tex]N =6.02\times 10^{23} (\frac{1}{2})^{\frac{20}{3.6}} [/tex]
[tex]N = 1.28 \times 10^{22}[/tex] atoms
So, after 20 days there will be [tex]1.28 \times 10^{22}[/tex] atoms of Pd-100.
Learn more about half-life in: brainly.com/question/24710827