# Coulomb's Law

A H nucleus & a nucleus of yet an unknown atom are stationary at 5.0x10^-10 m apart, Hydrogen atoms mass is mH = 1.7x10^-27kg with an electric charge of e = 1.6x10^-19 C. The Coulomb constant is Ke = 9.0x10^9 Nm^2 C^-2 and that the gravitational constant is G = 6.7x10^-11 N m^2 kg^-2.

Is the force between the hydrogen and the unknown atom is attractive or repulsive.

What is the electric charge of the unknown atom in coulombs with notation and as a whole number integer of the multiple of the charge of the H nucleus, e. Assumption is that the whole number is the atomic number of the unknown nucleus such as protons within it.

The magnitude of the gravitation force between the H nucleus and the unknowns is 9.3x10^-45 N. is this force attractive or repulsive. What is the mass of the unknown nucleus that is need to give a force of the magnitude in kg and notation. What is the mass as a whole number of a H nucleus mH with the assumption that this whole number is the mass number of the unknowns nucleus e.g. total number of nucleons.

And what is the isotope symbol for the unknown nucleus and what element does it belong to.

Â© BrainMass Inc. brainmass.com December 24, 2021, 7:19 pm ad1c9bdddfhttps://brainmass.com/physics/coulombs-law/force-hydrogen-nucleus-unknown-nucleus-179799

## SOLUTION This solution is **FREE** courtesy of BrainMass!

Please refer to the attachment.

A H nucleus & a nucleus of yet an unkown atom are stationary at 5.0x10-10 m apart. Hydrogen nucleus mass is mH = 1.7x10-27kg with an electric charge of e = 1.6x10-19 C. The Coulomb constant is ke = 9.0x109 Nm2 C-2 and that the gravitational constant is G = 6.7x10-11 N m2 kg-2.

The magnitude of electric force between the hydrogen nucleus and the unknown nucleus is 5.5x10-9 N. Is this force between the hydrogen and the unkown nucleus attractive or repulsive?

What is the electric charge on the unkown nucleus in coulombs expressed in scientific notation. Express this charge as a whole number multiple of the charge on the H nucleus, e. Assumption is that the whole number is the atomic number of the unkown nucleus i.e. the number of protons it contains.

The magnitude of the gravitation force between the H nucleus and the unkown nucleus is 9.3x10-45 N. Is this force attractive or repulsive ? What is the mass of the unkown nucleus that is needed to give a force of this magnitude in kg? What is the mass as a whole number multiple of the mass of a H nucleus i.e. mH, with the assumption that this whole number is the mass number of the unkowns nucleus i.e. total number of nucleons it contains.

What is the isotope symbol for the unkown nucleus and name the element.

Solution :

H X

mH = 1.7x10-27kg

e = +1.6x10-19 C r = 5x10-10m

The force between the two nuclei will be repulsive as both the nuclei are positively charged on account of presence of protons.

Let number of protons in the nucleus of X be Z. Thus, charge on the unknown nucleus = Ze.

As per Coulomb's law, electrostatic repulsive force between the two nuclei = kq1q2/r2 = ke(Ze)/r2 = kZe2/r2 = 9x109xZx(1.6x10-19)2/(5x10-10)2 = 0.922x10-9xZ Newton

Electrostatic force is given as 5.5x10-9 N.

Hence, 0.922x10-9xZ = 5.5x10-9 or Z = 6

Gravitational force = GmHmX/r2 = 6.7x10-11x1.7x10-27x mX/(5x10-10)2 = 0.4556x10-18 x mX N

where mX = mass of nucleus X

Gravitational force is given as 9.3x10-45 N

6.7x10-11x1.7x10-27x mX/(5x10-10)2 = 9.3x10-45

Or 0.4556x10-18 x mX = 9.3x10-45 or mX = 9.3x10-45/0.4556x10-18 = 20.41x10-27 kg

Atomic mass number of X = mX/mH = 20.41x10-27/1.7x10-27 = 12

The unknown nucleus has atomic number 6 and mass number 12. The isotope symbol of the unknown nucleus is 6C12 and the element is carbon.

Â© BrainMass Inc. brainmass.com December 24, 2021, 7:19 pm ad1c9bdddf>https://brainmass.com/physics/coulombs-law/force-hydrogen-nucleus-unknown-nucleus-179799