91Ó°ÊÓ

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

$\text{p=}\frac{\text{h}}{\text{λ}}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦(1)

$\text{E}=\text{hf}$……………….. (2)

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:

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

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

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

(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 $p=\sqrt{2mhf}$.

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:

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

$\text{p=}\frac{\text{h}}{\text{λ}}$

.

The equation for$E$is then as follows:

$\text{E}=\frac{{\text{h}}^{\text{2}}}{{\text{2³¾Î»}}^{\text{2}}}$

As a result, the energy-wavelength equation is $\text{E}=\frac{{\text{h}}^{\text{2}}}{{\text{2³¾Î»}}^{\text{2}}}$