Question

In the Bohr model of hydrogen atom, a single electron revolves a single proton in a circle of radius r = 410-11 m. a) By equating the electrical force to the electron mass times its acceleration, derive an expression for the electron’s speed and evaluate it. b) Obtain an expression for the kinetic energy and show that its magnitude is just half that of the electrical potential energy c) Obtain the expression for the total energy and evaluate it.

Answer 1

Answer:

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Answer 2

Given:

Bohr model of the hydrogen atom with orbital radius = r

To find:

a. Electron’s speed

b. Expression for the kinetic energy

c.  Expression for the total energy

Solution:

a. F= centripetal force = mv²/r

This force is provided by the coulomb's force =  k X Q² / r²

(Here, Q = e = 1.6 X 10⁻¹⁹ C and K = 9 X 10⁹ N m²C⁻²)

So mv²/r = k X Q² / r²

Solving for v (electron's speed) = $\sqrt{k e^2/ m r}$

b. Kinetic Energy = KE = 1/2 X m X v²

Substituting the value of v from (a),

KE = m X Ke²/ 2mr

= Ke²/2r

We know that PE = -k X e X e / r (PE of the electron is negative since the charge on the electron is negative)

This is just double the kinetic energy of the electron

c. E total = KE + PE

= Ke²/2r - Ke²/r

= -Ke²/r

The negative sign indicates that the electron is in a bounded state.