Derive Bohr's quantisation condition for angular momentum of orbiting electron in hydrogen atom using De Broglie's hypothesis.

To derive Bohr's quantization condition for the angular momentum of an orbiting electron in a hydrogen atom using De Broglie's hypothesis, we start by considering the electron's wave-like nature.

De Broglie's Hypothesis

According to De Broglie, particles such as electrons exhibit wave properties. The wavelength (\(\lambda\)) of an electron can be expressed as:

\(\lambda = \frac{h}{p}\)

Here, \(h\) is Planck's constant and \(p\) is the momentum of the electron, given by:

\(p = mv\)

where \(m\) is the mass of the electron and \(v\) is its velocity.

Orbital Condition

For an electron in a stable orbit around the nucleus, the circumference of the orbit must be an integer multiple of the wavelength:

\(2\pi r = n\lambda\)

where \(r\) is the radius of the orbit and \(n\) is a positive integer (the principal quantum number).

Substituting De Broglie's Wavelength

Now, substituting the expression for \(\lambda\) into the orbital condition gives:

2\pi r = n \left(\frac{h}{mv}\right)

Rearranging this, we find:

mv = \frac{nh}{2\pi r}

Angular Momentum Quantization

The angular momentum (\(L\)) of the electron is defined as:

L = mvr

Substituting the expression for \(mv\) from above into the angular momentum equation, we get:

L = r \left(\frac{nh}{2\pi r}\right) = \frac{nh}{2\pi}

Final Result

This leads us to Bohr's quantization condition for angular momentum:

L = n\frac{h}{2\pi}

Thus, the angular momentum of an electron in a hydrogen atom is quantized and can only take on discrete values based on the integer \(n\).