Question

Nonrelativistically, the energy$\mathit{E}$of a free massive particle is just kinetic energy, and its momentumis. of course,$\mathbf{\text{mv}}$. Combining these with fundamental relationships (4-4) and (4-5), derive a formula relating (a) particle momentumto matter-wave frequency fand (b) particle energy$\mathit{E}$to the wavelength$\mathbf{\text{λ}}$of a matter wave.

Short answer

a) The derivative formula by relating particle momentum to matter-wave frequency is $\text{p}=\sqrt{\text{2mhf}}$.

b) The derivative formula by relating particle energy to the wavelength of a matter wave is $\text{E}=\frac{{\text{h}}^{\text{2}}}{{\text{2mλ}}^{\text{2}}}$

Step-by-step solution

Relation of wave-particle momentum and energy.

The following equations describe the fundamental wave-particle momentum andenergy relationships, respectively.

$\text{p=}\frac{\text{h}}{\text{λ}}$…………………(1)

$\text{E}=\text{hf}$……………….. (2)

Relationship between momentum and frequency.

The relationship between momentum and frequency is described by an equation.

The energy can be expressed as$\text{E}=\frac{{\text{p}}^{\text{2}}}{\text{2m}}$

Putting this expression and equation$\left(\text{2}\right)$, we have:

$\text{E=}\frac{{\text{p}}^{\text{2}}}{\text{2m}}$

$\text{E}=\text{hf}$

$\frac{{\text{p}}^{\text{2}}}{\text{2m}}=\text{hf}$

Derivative equation for  

(a)

The equation for$p$is then as follows:

$\frac{{\text{p}}^{\text{2}}}{\text{2m}}=\text{hf}$

${\text{p}}^{\text{2}}=\text{2mhf}$

$p=\sqrt{2mhf}$

As a result, the equation for momentum and frequency is $p=\sqrt{2mhf}$.

Relation between energy and wavelength.

After that, we create an equation that connects energy and wavelength.

We know that,$\text{E=}\frac{{\text{p}}^{\text{2}}}{\text{2m}}$.

Now, relate this to the equation $\left(\text{1}\right)$, then the equation is as follows:

$\text{E=}\frac{{\text{p}}^{\text{2}}}{\text{2m}}$

$\text{p=}\frac{\text{h}}{\text{λ}}$

.

Derivative equation for energy-wavelength(b)

The equation for$E$is then as follows:

$\text{E}=\frac{{\text{h}}^{\text{2}}}{{\text{2mλ}}^{\text{2}}}$

As a result, the energy-wavelength equation is $\text{E}=\frac{{\text{h}}^{\text{2}}}{{\text{2mλ}}^{\text{2}}}$