91Ó°ÊÓ

Use the value the value of Plank’s constant as below.

$h=6.626×{10}^{−34}\text{J}â‹\dots \text{s}$

Mass of electron,$m=9.1×{10}^{−31}\text{kg}$

The energy,role="math" localid="1663130773211" $E=1\text{keV}={10}^3\text{ev}×1.6×{10}^{−19}\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{eV}$}\right.$

The wavelength that is associated with an object in relation to its momentum and mass is known as the de Broglie wavelength. The de Broglie wavelength of a particle is usually inversely proportional to its strength.

The de Broglie wavelength of the electron is,

$\begin{array}{rcl}\mathrm{λ}& =& \frac{h}{p}\\ & =& \frac{h}{mv}\end{array}$

Use the relativistic formula as follows:

$$\begin{gathered} E=\frac{1}{2}m{v}^2 \\ v=\sqrt{\frac{2E}{m}} \end{gathered}$$

Thus, wavelength of the electron is:

$\begin{array}{rcl}\mathrm{λ}& =& \frac{h}{\sqrt{2mE}}\\ & =& \frac{6.626×{10}^{−34}\text{J}â‹\dots \text{s}}{\sqrt{2×9.1×{10}^{−31}\text{kg}×({10}^3\text{ev}×1.6×{10}^{−19}\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{eV}$}\right.)}}\\ & =& 3.9×{10}^{−11}\text{m}\end{array}$

Hence, the wavelength of electron is $3.9×{10}^{−11}\text{m}$.

A photon’s de Broglie wavelength is equal to its familiar wave relationship value.

Use the value of$hc$as,

$hc=1240\text{​e³Õ}â‹\dots \text{nm}$

Therefore, de Broglie wavelength will becomes,

Hence, the wavelength of photon is $\begin{array}{rcl}& & 1.24\text{ â¶Ä‹n³¾}\end{array}$.

The neutron mass equals to $1.675×{10}^{−27}\text{kg}$.

Using the conversion from electron volts to Joules gives;

$\begin{array}{rcl}\mathrm{λ}& =& \frac{h}{p}\\ & =& \frac{h}{\sqrt{2m_{n}K}}\\ & =& \frac{hc}{\sqrt{2m_{n}eV}}\end{array}$

Substitute known values in the above equation.

$\begin{array}{rcl}\mathrm{λ}& =& \frac{6.626×{10}^{−34}\text{J}â‹\dots \text{s}}{\sqrt{2(1.67×{10}^{−27}\text{kg})(1.602×{10}^{−19}\text{C})\left({10}^3\text{V}\right)}}\\ & =& 2.65×{10}^{−4}\text{m}\end{array}$

Hence, the wavelength of the neutron is$2.65×{10}^{−4}\text{m}$.