Question

(a) What is the minimum potential difference between the filament and the target of an x-ray tube if the rube is to produce x rays with a wavelength of 0.150nm ? (b) What is the shortest wavelength produced in an x-ray tube operated at 30.0kV?

Short answer

(a) The minimum potential difference between the filament and the target of an x-ray tube is 8.27kV.

(b) The shortest wavelength produced in an x-ray tube is 0.014nm .

Step-by-step solution

Formula used to solve to question

$\mathit{e}{\mathit{V}}_{\mathbf{A}\mathbf{C}}\mathbf{=}\mathit{h}{\mathit{f}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\frac{\mathbf{h}\mathbf{c}}{\mathbf{\lambda }}$ (1)

Calculate the minimum potential difference

From equation (1),

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{V}}_{\mathbf{AC}}\mathbf{=}\frac{\mathbf{hc}}{\mathbf{e\lambda }} \\ \mathbf{\Rightarrow }{\mathit{V}}_{\mathbf{AC}}\mathbf{=}\frac{\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{34}}\mathbf{*}\mathbf{3}\mathbf{*}{\mathbf{10}}^{\mathbf{8}}}{\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{*}\mathbf{0}\mathbf{.}\mathbf{150}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}}\mathbf{=}\mathbf{8272}\mathbf{.}\mathbf{2}\mathit{V} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\Rightarrow }{\mathit{V}}_{\mathbf{AC}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{27}\mathit{k}\mathit{V} \end{gathered}$$

Calculate the shortest wavelength produced

From equation (1),

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{\lambda }}_{\mathbf{m}\mathbf{i}\mathbf{n}}\mathbf{=}\frac{\mathbf{h}\mathbf{c}}{\mathbf{e}{\mathbf{V}}_{\mathbf{A}\mathbf{C}}} \\ \mathbf{\Rightarrow }{\mathit{\lambda }}_{\mathbf{m}\mathbf{i}\mathbf{n}}\mathbf{=}\frac{\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{34}}\mathbf{*}\mathbf{3}\mathbf{*}{\mathbf{10}}^{\mathbf{8}}}{\mathbf{}\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{*}\mathbf{30}\mathbf{*}{\mathbf{10}}^{\mathbf{3}}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{14}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathit{m} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\Rightarrow }{\mathit{\lambda }}_{\mathbf{min}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{014}\mathit{n}\mathit{m} \end{gathered}$$

Thus, the minimum potential difference between the filament and the target of an x-ray tube is 8.27kV . The shortest wavelength produced in an x-ray tube is 0.014nm.