Question:medium

20 kV electrons can produce X-rays with a minimum wavelength of

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The shortcut formula \(\lambda (\text{nm}) \approx 1240/E(\text{eV})\) is extremely useful for problems involving photon energy and wavelength and is worth memorizing for competitive exams. Be mindful of the units (nm for 1240, Å for 12400).
Updated On: Mar 26, 2026
  • 0.248 Å
  • 0.41 Å
  • 0.099 nm
  • 0.062 nm
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The Correct Option is D

Solution and Explanation

Step 1: Formula for Cut-off Wavelength (Duane-Hunt Law). The minimum wavelength (\(\lambda_{min}\)) produced by an X-ray tube operating at voltage \(V\) is given by: \[ \lambda_{min} = \frac{hc}{eV} \] Using the shortcut value \(hc \approx 12400 \, \text{eV \AA} \) or \(1240 \, \text{eV nm}\).
Step 2: Calculation. Given Voltage \(V = 20 \, \text{kV} = 20,000 \, \text{V}\). Energy of electrons \(E = 20,000 \, \text{eV}\). Using \( \lambda (\text{nm}) \approx \frac{1240}{V (\text{volts})} \): Wait, the formula is \( \lambda \approx \frac{1240}{V} \text{ nm} \) if V is in Volts? No, \(E\) in eV. \[ \lambda_{min} = \frac{1240}{20000} \, \text{nm} \] \[ \lambda_{min} = \frac{124}{2000} \, \text{nm} = 0.062 \, \text{nm} \] Alternatively in Angstroms: \[ \lambda_{min} = \frac{12400}{20000} \, \text{\AA} = 0.62 \, \text{\AA} \]
Step 3: Check Options. Option (A): 0.248 \AA Option (B): 0.41 \AA Option (C): 0.099 nm Option (D): 0.062 nm (Matches our calculation)
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