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.
Electrons uniting with electrons of another atom is the cause in this relationship. The effect is a chemical change.
Variables we know:
t = 8 seconds
Vi = 0 m/s
g = -9.81
Δy = ?
Vf = ?
Equation we will be using to solve for Vf: Vf = Vi + gt
Steps to solve:
Vf = (0) + (-9.81)(8)
Vf = -78.48 m/s
Hope this helps!! :)
