Question

Plank's equation is E= hf f= E/h E-energy = ½mv², h- plank's constant, f - frequency. λf = v λ= v/ f = vh/ E = 2h/p P- momentum = mv

but de broglie equation is λ = h/p. Why?

Answer 1

Answer on Question #67227 - Physics – Quantum Mechanics

Question:

Plank's equation is $E = hf f = E/h$ E-energy $= \frac{1}{2}mv^2$, h- plank's constant, f - frequency. $\lambda f = v \lambda = v/f = vh/E = 2h/p$ P- momentum $= mv$

but de Broglie equation is $\lambda = h/p$. Why?

Solution:

$$\lambda = h/mV$$

It is necessary to note the dishonest reception of official physics, which she often uses to camouflage her own contradictions. In formula (1), by the Planck constant $h$ we mean a completely definite numerical value, depending only on the choice of the system of units of measure, but not dependent on a specific physical object. To bring this fraud to clean water, substitute in (1) Planck's constant in another form: $= h / 2p$:

$L = 2p / mV$ (2), but $l / 2p = r$, where $r$ is the radius of the helical trajectory, then (2) becomes the particle momentum momentum definition:

$= MVr$ (3). But according to the ideas of official physics, the "bosons" have a whole or zero, and the "fermions" have a half-integer, although both obey the de Broglie formula (1); Actually correspond to formula (3). Consequently, neither "bosons" nor "fermions" exist in nature. They exist only in the minds of theorists.

Another gross mistake of official physics is that the enormous mass of macrobodies is substituted into the denominator (1), they receive an astonishingly small "wavelength" of these bodies and thereby "confirm" the correctness (1). But then, if you want to use the numerator to substitute a huge moment of macrobodies, then you will get a "wavelength" of cosmic scales.

The new physics went farther than de Broglie and asserts corpuscular-wave dualism for any free bodies, including macrobodies.

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