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Q16E

Expert-verifiedFound in: Page 134

Book edition
2nd Edition

Author(s)
Randy Harris

Pages
633 pages

ISBN
9780805303087

**A brag diffraction experiment is conducted using a beam of electrons accelerated through a${\mathbf{\text{1.0\hspace{0.17em}kV}}}$ ****potential difference. (a) If the spacing between atomic planes in the crystal is ${\mathbf{\text{0.1 nm}}}$**** , at what angles with respect to the planes will diffraction maximum be observed? (b) If a beam of ${\mathbf{X}}$****-rays products diffraction maxima at the same angles as the electron beam, what is the${\mathbf{X}}$ ****-ray photon energy?**

(a) The angles for the maximum is

$\begin{array}{l}{\text{\theta}}_{\text{1}}=\text{11.2}\xb0\\ {\text{\theta}}_{\text{2}}=\text{22.83}\xb0\\ {\text{\theta}}_{\text{3}}=\text{35.59}\xb0\text{}\\ {\text{\theta}}_{\text{4}}=\text{50.89}\xb0\text{}\\ {\text{\theta}}_{\text{5}}=\text{75.93}\xb0\end{array}$

(b) The photon energy of the x-ray is

${\text{E}}_{\text{p}}=\text{31.88 keV}$

**So, we'll use the relationship shown in Example ${\mathbf{\text{4.3}}}$directly, linking the particle's wavelength ${\mathbf{\text{(}}}{\mathit{\lambda}}{\mathbf{\text{)}}}$to the potential for rapid growth (V). can use Bragg's law to find the equivalent values of once have the wavelength, such that,**

**${\mathit{\lambda}}{\mathbf{\text{=}}}\sqrt{\frac{{\mathbf{\text{h}}}^{\mathbf{\text{2}}}}{\mathbf{\text{2mqV}}}}$………………..(1)**

Substitute the given data in equation (1), and we get,

$\lambda =\sqrt{\frac{{(\text{6.63}\times {\text{10}}^{\text{-34}}\text{\hspace{0.17em}}J\cdot s)}^{\text{2}}}{\text{2}\times \text{9.1}\times {\text{10}}^{\text{-31}}\text{\hspace{0.17em}}kg\times \text{1.6}\times {\text{10}}^{\text{-19}}\text{\hspace{0.17em}C}\times \text{1000 V}}}$$\begin{array}{c}\lambda =\sqrt{\text{1.5}\times {\text{10}}^{\text{-21}}{\text{m}}^{\text{2}}}\\ =\text{3.89}\times {\text{10}}^{\text{-11}}\text{m}\end{array}$

$\begin{array}{c}\text{2dsin(\theta )}=\text{n}\lambda \\ \text{sin(\theta )}=\left(\frac{\lambda}{\text{2d}}\right)\text{n}\\ \text{sin(\theta )}=\left(\frac{\text{3.89}\times {\text{10}}^{\text{-11}}\text{m}}{\text{2}\times \text{0.1}\times {\text{10}}^{\text{-9}}\text{m}}\right)\times \text{n}\\ \text{\theta}={\text{sin}}^{\text{-1}}\left(\text{0.194n}\right)\end{array}$

$\begin{array}{l}{\text{\theta}}_{\text{1}}=\text{11.2}\xb0\\ {\text{\theta}}_{\text{2}}=\text{22.83}\xb0\\ {\text{\theta}}_{\text{3}}=\text{35.59}\xb0\text{}\\ {\text{\theta}}_{\text{4}}=\text{50.89}\xb0\text{}\\ {\text{\theta}}_{\text{5}}=\text{75.93}\xb0\end{array}$

Keep in mind that the value of$\text{n=6}$ isn't allowed because arcsine's parameter exceeds$\text{1}$ , but used the integer n instead of m for the integer counter so it doesn't get mixed up with the unit of measurement ($\text{m}$ or meter).

(b)

**To achieve the same maxima, both the electron and the photons must have the same wavelength value.**

${\text{E}}_{\text{p}}\text{=}\frac{\text{hc}}{\lambda}$

$\begin{array}{c}{\text{E}}_{\text{p}}=\left(\frac{4.14\times {10}^{-23}\text{\hspace{0.17em}}eV\cdot s\times \text{3}\times {\text{10}}^{\text{8}}\text{\hspace{0.17em}m}/s}{\text{3.89}\times {\text{10}}^{\text{-11}}\text{m}}\right)\text{\hspace{0.17em}}\\ =\text{31.88 keV}\end{array}$

Therefore, the energy of the photon is${\text{E}}_{\text{p}}=\text{31.88 keV}$ .

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