What are (a) the energy of a photon corresponding to wavelength $\mathbf{1}\mathbf{.00}\mathbf{\text{nm}}$, (b) The kinetic energy of an electron with de Broglie wavelength $\mathbf{1}\mathbf{.00}\mathbf{\text{nm}}$, (c) the energy of a photon corresponding to wavelength$\mathbf{1}\mathbf{.00}\mathbf{\text{fm}}$, and (d) the kinetic energy of an electron with de Broglie wavelength$\mathbf{1}\mathbf{.00}\mathbf{\text{fm}}$?

Different de Broglie wavelengths are given. The energy corresponding to these wavelengths of photon and electron is demanded.

The energy of the photon is,

$\mathit{E}\mathbf{=}\frac{\mathbf{h}\mathbf{c}}{\mathbf{\lambda }}$

Here, $\mathbf{h}$ is Plank’s constant, $\mathbf{c}$is the speed of light, and$\mathbf{\lambda }$is the wavelength.

The value of a constant $hc$is,

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

The kinetic energy of the electron is,

role="math" localid="1663146719321" $\begin{array}{rcl}\mathit{K}& \mathbf{=}& \frac{{\mathbf{p}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{m}}_{\mathbf{e}}}\\ & \mathbf{=}& \frac{{\mathbf{\left(}{\displaystyle \raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$\lambda $}\right.}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{m}}_{\mathbf{e}}}\\ & \mathbf{=}& \frac{{\mathbf{\left(}{\displaystyle \raisebox{1ex}{$hc$}\!\left/ \!\raisebox{-1ex}{$\lambda $}\right.}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{m}}_{\mathbf{e}}{\mathbf{c}}^{\mathbf{2}}}\end{array}$

Here, $\begin{array}{rcl}& & \mathbf{p}\end{array}$is momentum and $\begin{array}{rcl}& & {\mathbf{m}}_{\mathbf{e}}\end{array}$is the mass of the electron.

Given wavelength,

$\lambda =1\text{nm}$

Write the equation for energy as,

$E=\frac{hc}{\lambda }$

Substitute $1240\text{nm}\cdot \text{eV}$ for $hc$ and $\mathrm{1\hspace{0.17em}}\mathrm{nm}$ for role="math" localid="1663146920122" $\lambda$ in the above equation.

$\begin{array}{rcl}E& =& \frac{\text{1240nm}\cdot \text{eV}}{\text{1\hspace{0.17em}nm}}\\ & =& 1240\text{eV}\left(\frac{1keV}{1000eV}\right)\\ & =& 1.24\text{\hspace{0.17em}keV}\end{array}$

Hence, energy of a photon corresponding to wavelength $\text{1\hspace{0.17em}nm}$is $1.24\text{\hspace{0.17em}keV}$.

The energy of the electron is,

$K=\frac{{\left({\displaystyle \raisebox{1ex}{$hc$}\!\left/ \!\raisebox{-1ex}{$\lambda $}\right.}\right)}^2}{2m_e{c}^2}$

Given de Broglie wavelength

$\lambda =1\text{\hspace{0.17em}nm}$

So the kinetic energy will be,

$\begin{array}{rcl}K& =& \frac{{\left({\displaystyle \frac{1240\text{\hspace{0.17em}nm}\cdot \text{eV}}{1\text{\hspace{0.17em}nm}}}\right)}^2}{2\left(0.511\text{\hspace{0.17em}MeV}\right)}\\ & =& 1.50\text{eV}\end{array}$

Hence, energy of an electron corresponding to wavelength $1\text{\hspace{0.17em}nm}$is$1.50\text{eV}$.

Given wavelength

$\lambda =\text{1\hspace{0.17em}fm}\left(\frac{{\text{10}}^{\text{-6}}\text{\hspace{0.17em}nm}}{\text{1\hspace{0.17em}fm}}\right)=1\times {10}^{-6}\text{\hspace{0.17em}nm}$

Write the equation for energy as below.

$E=\frac{hc}{\lambda }$

Substitute $1240\text{nm}\cdot \text{eV}$for $hc$and $1\times {10}^{-6}\text{nm}$for $\lambda$in the above equation.

$\begin{array}{rcl}E& =& \frac{1240\text{\hspace{0.17em}nm}\cdot \text{eV}}{1\times {10}^{-6}\text{nm}}\\ & =& 1.24\times {10}^9\text{\hspace{0.17em}eV}\left(\frac{1\text{\hspace{0.17em}GeV}}{{10}^9\text{\hspace{0.17em}eV}}\right)\\ & =& 1.24\text{GeV}\end{array}$

Hence, energy of photon corresponding to wavelength$1\text{\hspace{0.17em}fm}$ is $1.24\text{GeV}$.

The kinetic energy of the electron is,

$K=\sqrt{{\left(\frac{hc}{\lambda }\right)}^2+m_e^2{c}^4}-m_e{c}^2$ ….. (1)

Here, $c$is the speed of light, $\lambda$is wavelength, and $m_e$is the mass of the electron.

Given de Broglie wavelength as,

$\lambda =1\text{\hspace{0.17em}fm}$

The constant, $hc=1240\text{\hspace{0.17em}MeV}\cdot \text{fm}$

And the constant, $m_e{c}^2=0.511\text{\hspace{0.17em}MeV}$

Substitute these values into equation (1).

$\begin{array}{rcl}K& =& \sqrt{{\left(\frac{1240\text{\hspace{0.17em}nm}\cdot \text{eV}}{1\text{\hspace{0.17em}nm}}\right)}^2+{\left(0.511\text{\hspace{0.17em}MeV}\right)}^2}-0.511\text{\hspace{0.17em}MeV}\\ & =& 1.24\times {10}^3\text{MeV}\left(\frac{\text{1\hspace{0.17em}GeV}}{{\text{10}}^{\text{3}}\text{\hspace{0.17em}MeV}}\right)\\ & =& 1.24\text{GeV}\end{array}$

Hence, energy of the electron corresponding to wavelength $1\text{\hspace{0.17em}fm}$is $\begin{array}{rcl}& & 1.24\text{GeV}\end{array}$.