The velocity of electron in hydrogen atom is 7.29 × 10⁷ cm/sec. The potential energy of that electron is

It is a well-known fact in physics that the kinetic energy (K.E.) of an electron in a hydrogen atom is related to its potential energy (P.E.) as:

Kinetic Energy = - (Potential Energy) / 2

Formula for Kinetic Energy:

Kinetic Energy is given by the expression:

K.E. = (m × V²) / 2

Here:

m = Mass of the electron (9.109 × 10^-31 kg)

V = Velocity of electron (7.29 × 10⁷ cm/sec or 7.29 × 10⁵ m/sec)

Substitute Values:

The velocity of electron is converted to meters per second for consistency. Substituting the values into the formula:

K.E. = (9.109 × 10^-31 × (7.29 × 10⁵)²) / 2

Simplifying this:

K.E. = 2.418 × 10^-19 J

Determine Potential Energy:

Using the relation K.E. = -(P.E.) / 2, we find that:

P.E. = -2 × K.E.

Substituting the value of K.E.:

P.E. = -2 × 2.418 × 10^-19 J

P.E. = -4.836 × 10^-19 J

Convert to Electron Volts:

To express the potential energy in electron volts (eV), use the conversion factor:

1 eV = 1.602 × 10^-19 J

Thus:

P.E. = (-4.836 × 10^-19 J) / (1.602 × 10^-19 J/eV)

P.E. = -3.02 eV

Final Answer

The potential energy of the electron, based on the given velocity of electron, is approximately:

P.E. = -4.836 × 10^-19 J or -3.02 eV

This detailed calculation demonstrates the relationship between the velocity of electron, its kinetic energy, and potential energy in a hydrogen atom.