Question

What electron energy is required to obtain the diffraction pattern for a surface with crystal spacing of $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{°}\mathbf{\text{A}}$.

Short answer

The energy required for an electron beam is $4.96\times {10}^{-16}J$.

Step-by-step solution

Diffraction

Diffraction refers to a phenomenon that takes place when electromagnetic radiation comes in contact with an obstacle. With this, the interference and bending of waves occur.

Calculation

The wavelength of the wave with the angle of diffraction and crystal spacing is given as:

$\lambda =a\sin \theta$

The maximum possible value of wavelength will be as:

$\begin{array}{rcl}\lambda & =& a\sin 90°\\ & =& a\end{array}$

Here the value of angstrom is converted to m as:

$\begin{array}{rcl}1°A& =& {10}^{-10}m\\ 4°A& =& 4\times {10}^{-10}m\end{array}$

The energy associated with the electronic transition is given by the formula as:

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

*E is the energy of the electronic wave.

*h is the Planck’s; constant, i.e.,$6.6.25\times {10}^{-34}Js$.

*c is the velocity of light, i.e.,$3\times {10}^{8}m{s}^{-1}$.

*is the wavelength of the associated wave, i.e.,$4\times {10}^{-10}m$.

On substituting all the given values, we get

$\begin{array}{rcl}E& =& \frac{hc}{\lambda }\\ & =& \frac{6.6.25\times {10}^{-34}Js\times 3\times {10}^{8}m{s}^{-1}}{4\times {10}^{-10}m}\\ & =& 4.96\times {10}^{-16}J\\ & & \end{array}$

Thus, the energy required for an electron beam is $4.96\times {10}^{-16}J$.