91Ó°ÊÓ

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

The energy of the photon is,

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

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

The value of a constant $hc$is,

$hc=1240\text{nm}â‹\dots \text{eV}$

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$}\!\left/ \!\raisebox{-1ex}{$\mathrm{λ}$}\right.}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{m}}_{\mathbf{e}}}\\ & \mathbf{=}& \frac{{\mathbf{\left(}{\displaystyle \raisebox{1ex}{$hc$}\!\left/ \!\raisebox{-1ex}{$\mathrm{λ}$}\right.}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{m}}_{\mathbf{e}}{\mathbf{c}}^{\mathbf{2}}}\end{array}$

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,

$\mathrm{λ}=1\text{nm}$

Write the equation for energy as,

$E=\frac{hc}{\mathrm{λ}}$

Substitute $1240\text{nm}â‹\dots \text{eV}$ for $hc$ and $\mathrm{1 }\mathrm{nm}$ for role="math" localid="1663146920122" $\mathrm{λ}$ in the above equation.

$\begin{array}{rcl}E& =& \frac{\text{1240nm}â‹\dots \text{eV}}{\text{1 n³¾}}\\ & =& 1240\text{eV}\left(\frac{1keV}{1000\text{ }eV}\right)\\ & =& 1.24\text{ k±ð³Õ}\end{array}$

Hence, energy of a photon corresponding to wavelength $\text{1 n³¾}$is $1.24\text{ k±ð³Õ}$.

The energy of the electron is,

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

Given de Broglie wavelength

$\mathrm{λ}=1\text{ n³¾}$

So the kinetic energy will be,

$\begin{array}{rcl}K& =& \frac{{\left({\displaystyle \frac{1240\text{ n³¾}â‹\dots \text{eV}}{1\text{ n³¾}}}\right)}^2}{2\left(0.511\text{ M±ð³Õ}\right)}\\ & =& 1.50\text{eV}\end{array}$

Hence, energy of an electron corresponding to wavelength $1\text{ n³¾}$is$1.50\text{eV}$.

Given wavelength

$\mathrm{λ}=\text{1 f³¾}\left(\frac{{\text{10}}^{\text{-6}}\text{ n³¾}}{\text{1 f³¾}}\right)=1×{10}^{−6}\text{ n³¾}$

Write the equation for energy as below.

$E=\frac{hc}{\mathrm{λ}}$

Substitute $1240\text{nm}â‹\dots \text{eV}$for $hc$and $1×{10}^{−6}\text{nm}$for $\mathrm{λ}$in the above equation.

$\begin{array}{rcl}E& =& \frac{1240\text{ n³¾}â‹\dots \text{eV}}{1×{10}^{−6}\text{nm}}\\ & =& 1.24×{10}^9\text{ e³Õ}\left(\frac{1\text{ G±ð³Õ}}{{10}^9\text{ e³Õ}}\right)\\ & =& 1.24\text{GeV}\end{array}$

Hence, energy of photon corresponding to wavelength$1\text{ f³¾}$ is $1.24\text{GeV}$.

The kinetic energy of the electron is,

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

Here, $c$is the speed of light, $\mathrm{λ}$is wavelength, and $m_e$is the mass of the electron.

Given de Broglie wavelength as,

$\mathrm{λ}=1\text{ f³¾}$

The constant, $hc=1240\text{ M±ð³Õ}â‹\dots \text{fm}$

And the constant, $m_e{c}^2=0.511\text{ M±ð³Õ}$

Substitute these values into equation (1).

$\begin{array}{rcl}K& =& \sqrt{{\left(\frac{1240\text{ n³¾}â‹\dots \text{eV}}{1\text{ n³¾}}\right)}^2+{\left(0.511\text{ M±ð³Õ}\right)}^2}−0.511\text{ M±ð³Õ}\\ & =& 1.24×{10}^3\text{MeV}\left(\frac{\text{1 G±ð³Õ}}{{\text{10}}^{\text{3}}\text{ M±ð³Õ}}\right)\\ & =& 1.24\text{GeV}\end{array}$

Hence, energy of the electron corresponding to wavelength $1\text{ f³¾}$is $\begin{array}{rcl}& & 1.24\text{GeV}\end{array}$.