Chapter 27, Problem 66GP

To determine

The spacing between planes in the diffracting crystal.

Answer to Problem 66GP

  $0.108nm$

Explanation of Solution

Given info:

The kinetic energy is $\text{85}\text{eV}$ and angle where first order peak is observed at $\text{θ}\text{=}\text{38°}$ .

Formula used:

Each absorbed photon’s momentum changes form $\text{p}\text{=}\frac{\text{h}}{\text{λ}}$

  $\lambda$ is wavelength.

  $p$ is photon’s momentum

  $h$ is Planck’s constant, $6.63\times {10}^{-34}J\cdot s$

It is known that $\sin \theta =\frac{m\lambda }{d}$

Where,

  $m$ is an integer

  $\lambda$ is wavelength

  $d$ is a distance

Calculation:

Because the energy is much less than ${\text{m}}_{\text{0}}{\text{c}}^{\text{2}}$ , use the formula of kinetic energy, $\text{KE}\text{=}\frac{{\text{p}}^{\text{2}}}{{\text{2m}}_{\text{0}}}$ .

Now, the wavelength of the electron:

  $\text{λ = }\frac{\text{h}}{\text{p}}\text{ = }\frac{\text{h}}{{\left[{\text{2m}}_{\text{0}}\left(\text{KE}\right)\right]}^{\frac{\text{1}}{\text{2}}}}\text{ = }\frac{\text{hc}}{{\left[{\text{2m}}_{\text{0}}{\text{c}}^{\text{2}}\left(\text{KE}\right)\right]}^{\frac{\text{1}}{\text{2}}}}$

Where, $c$ is speed of light, ${\text{3×10}}^{\text{8}}\text{ m/s }$

Substitute all the given values in the formula:

  $\text{ = }\frac{\left(\text{1}{\text{.24×10}}^{\text{3}}\text{eV}\text{×}\text{nm}\right)}{{\left[\text{2}\left(\text{0}{\text{.511×10}}^{\text{6}}\text{eV}\right)\left(\text{85}\text{eV}\right)\right]}^{\frac{\text{1}}{\text{2}}}}\text{=}\text{0}\text{.133}\text{nm}$

Now, calculate the spacing of the planes form the formula $2d\sin \theta =m\lambda$

  $2d\sin 38°=\left(1\right)\left(0.133nm\right)$

This gives $d=0.108nm$

Conclusion:

The required value is $0.108nm$ .