What potential difference must be applied to an x-ray tube to produce x-rays with a minimum wavelength of \(45.0 \mathrm{pm} ?\)

Short Answer

Expert verified
Answer: The required potential difference is approximately 2.94 x 10^4 V.

Step by step solution

01

Convert the given wavelength from pm to meters

First, we need to convert the given minimum wavelength from picometers to meters: \(\lambda_\text{min} = 45.0 \mathrm{pm} \times \dfrac{1 \mathrm{m}}{10^{12} \mathrm{pm}} = 45.0 \times 10^{-12} \mathrm{m}\)
02

Rearrange the Duane-Hunt law to solve for V

Next, we rearrange the Duane-Hunt law to isolate \(V\) on one side of the equation: \(V = \dfrac{hc}{e\lambda_\text{min}}\)
03

Substitute the known values and calculate V

Now, we substitute the known values of \(h\), \(c\), \(e\), and \(\lambda_\text{min}\) into the equation and calculate \(V\): \(V = \dfrac{(6.6261 \times 10^{-34} \mathrm{J\cdot s})(2.9979 \times 10^8 \mathrm{m/s})}{(1.6022 \times 10^{-19} \mathrm{C})(45.0 \times 10^{-12} \mathrm{m})}\) \(V \approx 2.94 \times 10^4 \mathrm{V}\) The required potential difference to produce x-rays with a minimum wavelength of \(45.0 \mathrm{pm}\) is approximately \(2.94 \times 10^4 \mathrm{V}\).

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