Question

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

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

=====

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

Answer 2

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:

                                            $N =N_{0} (\frac{1}{2})^{t/t_{1/2}$

Since the initial amount of mass is equal to $6.02\times10^{23}$ 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:

                                         $N =6.02\times 10^{23} (\frac{1}{2})^{\frac{20}{3.6}}$

                                            $N = 1.28 \times 10^{22}$ atoms

So, after 20 days there will be $1.28 \times 10^{22}$ atoms of Pd-100.

Learn more about half-life in: brainly.com/question/24710827