Question

Calculate the de Broglie wavelength of (a) a $\mathbf{\text{1.00keV}}$ electron, (b) a$\mathbf{\text{1.00keV}}$ photon, and (c) a$\mathbf{\text{1.00keV}}$neutron.

Short answer

(a) The wavelength of an electron is $3.9\times {10}^{-11}\text{m}$.

(b) The wavelength of the photon is $1.24\text{\hspace{0.17em}nm}$.

(c) The wavelength of the neutron is $2.65\times {10}^{-4}\text{m}$.

Step-by-step solution

The known data:

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

$h=6.626\times {10}^{-34}\text{J}\cdot \text{s}$

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

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

A concept of wavelength:

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.

(a) The wavelength of a 1.00 keV  electron:

The de Broglie wavelength of the electron is,

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

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:

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

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

(b) The wavelength of a 1.00 keV photon:

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

Use the value of$hc$as,

$hc=1240\text{eV}\cdot \text{nm}$

Therefore, de Broglie wavelength will becomes,

$\begin{array}{rcl}\lambda & =& \frac{hc}{E}\\ & =& \frac{1240\text{eV}\cdot \text{nm}}{1.00\text{\hspace{0.17em}keV}}\\ & =& 1.24\text{\hspace{0.17em}nm}\end{array}$

Hence, the wavelength of photon is $\begin{array}{rcl}& & 1.24\text{\hspace{0.17em}nm}\end{array}$.

(c) The wavelength of a 1.00 keV neutron:

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

Using the conversion from electron volts to Joules gives;

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

Substitute known values in the above equation.

$\begin{array}{rcl}\lambda & =& \frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}}{\sqrt{2(1.67\times {10}^{-27}\text{kg})(1.602\times {10}^{-19}\text{C})\left({10}^3\text{V}\right)}}\\ & =& 2.65\times {10}^{-4}\text{m}\end{array}$

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