For momentum per particle, given a 2D particle of area \(\pi a^2_\psi\),
\(2\pi a_{ \psi }=n\lambda \)
\(p= p_\rho\pi a^2_\psi=m_{ \rho }c.\pi a^{ 2 }_{ \psi }=mc\)
As defined in the previous post "de Broglie Per Unit Volume",
\( h_{ \rho }=2\pi a_{ \psi }m_{ \rho }c\)
as we consider per particle,
\( h=h_{ \rho }\pi a^{ 2 }_{ \psi }\)
\( h=2\pi a_{ \psi }mc=mc.n\lambda \)
So,
\( p=\cfrac { h }{ n\lambda } \)
when \(n=1\) we have,
\( p=\cfrac { h }{ \lambda } \)
which is simply obtained by multiplying the dimensions (volume, area or length) of the particle to both sides of the momentum equation derived in the post "de Broglie Per Unit Volume".
\( p_\rho.D=\cfrac { h_\rho }{ n\lambda }.D \)
where \(D=\pi a^2\) (2D particle) or \(D=a\) (1D particle), is the physical dimension of the particle. When we set \(n=1\), we have the same result.
But why does \(p\) decreases with \(n\) when \(n\ne 1\)?
Consider,
\(pc=\cfrac { h }{ n } f=mc^2\)
So,
\(E=\cfrac{1}{n}hf\)
Not \(E=nhf\).
Obviously when \(n=1\),
\(E=hf\)
In this derivation \(n\) wavelengths are packed into a fixed perimeter of \(2\pi a_\psi\). This is different from the normal treatment of \(n\), where \(a_\psi\) is allowed to change as \(n\) is incremented progressively for valid values of \(\lambda\). In the latter case, \(\lambda\) increases with \(n\). In the former case, \(\lambda\) decreases with \(n\). Both treatments are equivalent when \(n=1\).
Both treatment provides valid answers to \(\lambda\). Maybe it is better to name the number of wavelengths packed into a fixed perimeter of \(2\pi a_\psi\) as \(n_a\).
So,
\( p=\cfrac{1}{n_a}\cfrac { h }{ \lambda } \) and \(E=\cfrac{1}{n_a}hf\)
where
\( h=2\pi a_{ \psi }mc\),
\(m\) the mass of the particle, and \(a_\psi\) the extend of \(\psi\) around the particle.
We resolved the inconsistency and derived the Planck's constant. Since \(m_\rho\) is the same for all particles,
\(m=m_\rho.D\)
where \(D\) is either \(\pi a^2\) or \(a\), \(a\) is the physical dimension of the particle.
\(h\) is a constant for a particular type of particle but not across all particles. In the case of
\(h_\rho=2\pi a_{ \psi }m_{\rho}c\)
this expression depends on \(a_\psi\) only, as \(m_\rho\) is a constant.
Note: This derivation is not of a particle in orbit around any nucleus. This is just the particle itself with an energy density, \(\psi\) around it.
