Question

A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this atom? rnucleus = _____ m b. What is the magnitude of the electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus. F=_______ N

Answer 1

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Step-by-step explanation:

`r=r_o* A^{(1)/(3)}`

`r_o=1.25 * 10^(-15) m` = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

`r=1.25 * 10^(-15) m* (52)^{(1)/(3)}`

`r=4.6656* 10^(-15) m=4.6656 fm`

The approximate radius of the nucleus of this atom is 4.656 fermi.

b)
`F=k* (q_1q_2)/(a^2)`

k=
`9* 10^9 N m^2/C^2` = Coulombs constant

`q_1,q_2` = charges kept at distance 'a' from each other

F = electrostatic force between charges

`q_1=+1.602* 10^(-19) C`

`q_2=+1.602* 10^(-19) C`

Force of repulsion between two protons on opposite sides of the diameter

`a=2* r=2* 4.6656* 10^(-15) m=9.3312* 10^(-15) m`

`F=9* 10^9 N m^2/C^2* ((+1.602* 10^(-19) C)* (+1.602* 10^(-19) C))/((9.3312* 10^(-15) m)^2)`

`F=2.6527 N`

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527