Respuesta :
Answer:
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
Explanation:
[tex]r=r_o\times A^{\frac{1}{3}}[/tex]
[tex]r_o=1.25 \times 10^{-15} m[/tex] = 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
[tex]r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}[/tex]
[tex]r=4.6656\times 10^{-15} m=4.6656 fm[/tex]
The approximate radius of the nucleus of this atom is 4.656 fermi.
b) [tex]F=k\times \frac{q_1q_2}{a^2}[/tex]
k=[tex]9\times 10^9 N m^2/C^2[/tex] = Coulombs constant
[tex]q_1,q_2[/tex] = charges kept at distance 'a' from each other
F = electrostatic force between charges
[tex]q_1=+1.602\times 10^{-19} C[/tex]
[tex]q_2=+1.602\times 10^{-19} C[/tex]
Force of repulsion between two protons on opposite sides of the diameter
[tex]a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m[/tex]
[tex]F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}[/tex]
[tex]F=2.6527 N[/tex]
The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527
