The radioactive decay obeys first order kinetics
the rate law expression for radioactive decay is
![ln\frac{[A_{0}]}{[A_{t}]}=kt](https://tex.z-dn.net/?f=ln%5Cfrac%7B%5BA_%7B0%7D%5D%7D%7B%5BA_%7Bt%7D%5D%7D%3Dkt)
Where
A0 = initial concentration
At = concentration after time "t"
t = time
k = rate constant
For first order reaction the relation between rate constant and half life is:
Let us calculate k
k = 0.693 / 72 = 0.009625 years⁻¹
Given
At = 0.25 A0
time = 144 years
So after 144 years the sample contains 25% parent isotope and 75% daughter isotopes**
Simply two half lives
The particles cannot move around at all. The particles are, however, still in motion.
Answer:
Yea.....where is the article and the 6 questions?
Answer:
This question is incomplete but the correct option is B
Explanation:
This question is incomplete because of the absence of the "Reference Table S", however the question can still be answered in the absence of the table. The energy described in the question is the ionization energy (energy required to remove the most loosely bound electron in an atom). This question seeks to know the atom (from the options provided) with the least ionization energy.
Ionization energy increases from left to right across the period because it's easier to remove a single electron (valence electron) from the outermost shell than to remove two electrons from the same shell; thus the more the valence electrons (in a shell), the higher the ionization energy. Thus, bromine (Br) and tin (Sn) have high ionization energies because they have more number of electrons in there outermost shell.
<u>Berylium (Be) and strontium (Sr) are both in the group 2 of the periodic table because they both have 2 electrons in there outermost shell. Ionization energy decreases down a group. This is because the farther an electron is from the nucleus, the weaker the force of attraction between the nucleus and the electron. Thus, strontium (Sr) would have a lesser ionization energy between the two and would indeed have the least ionization among the options provided</u>. Hence, the correct option is B
