They are called Cargo Shorts.
Answer:
Take E(alpha particle energy) = 5.5 MeV (5.5x106x1.6x10-19)
If the charge on the lead nucleus is +82e(atomic number of lead is 82) = +82x1.6x10-19 C and the charge on the alpha particle is +2e = 2x1.6x10-19 C
Using dc = (1/4πεo)qQ/Eα we have
dc = [9x10^9x(2x1.6x10-19x82x1.6x10-19)]/5.5x10-13 = 6.67x10^-13m. = 6.67 x 10^-13 x 10^15 = 6.67 x 10^2fm
Note: 1meter = 10^15fentometer
Explanation:
This is well inside the atom but some eight nuclear diameters from the centre of the lead nucleus.
The answer is: lose electrons and form positive ions.
Most metals have strong metallic bond, because of strong electrostatic attractive force between valence electrons (metals usually have low ionization energy and lose electrons easy) and positively charged metal ions.
The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).
For example, magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.
Electron configuration of magnesium ion: ₁₂Mg²⁺ 1s² 2s² 2p⁶.
I think the amount would be a 0.4998 mol
I did moles=mass(g)/A,r
=12.5/24.3 to get that
The correct answer is slow
