Question

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

• A. 0.248 Å
• B. 0.41 Å
• C. 0.099 nm
• D. 0.062 nm

Hint:

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).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The production of X-rays in an X-ray tube involves accelerating electrons through a high potential difference and making them strike a metal target. The minimum wavelength (or maximum frequency) of the produced X-rays corresponds to the scenario where an electron loses all its kinetic energy in a single collision, converting it into a single X-ray photon. This is related to the cutoff wavelength of the continuous X-ray spectrum.
Step 2: Key Formula or Approach:
The kinetic energy (\(K\)) gained by an electron accelerated through a potential difference \(V\) is \(K = eV\).
The energy (\(E\)) of a photon of wavelength \(\lambda\) is \(E = \frac{hc}{\lambda}\).
For the minimum wavelength (\(\lambda_{min}\)), the maximum possible photon energy is equal to the electron's kinetic energy:
\[ eV = \frac{hc}{\lambda_{min}} \] Rearranging for \(\lambda_{min}\):
\[ \lambda_{min} = \frac{hc}{eV} \] A very useful approximation for calculations is:
\[ \lambda_{min} \approx \frac{1240 \text{ eV}\cdot\text{nm}}{e \cdot V} = \frac{1240}{V} \text{ nm} \] where V is the potential difference in Volts.
Step 3: Detailed Explanation:
Given values:
- Accelerating potential, \(V = 20 \text{ kV} = 20 \times 10^3 \text{ V}\).
Using the shortcut formula:
\[ \lambda_{min} (\text{in nm}) = \frac{1240}{V (\text{in Volts})} \] \[ \lambda_{min} = \frac{1240}{20 \times 10^3} = \frac{124}{2 \times 10^3} = \frac{62}{1000} = 0.062 \text{ nm} \] Let's check the answer in Angstroms (Å) as well, since some options are in Å. (1 nm = 10 Å).
\[ \lambda_{min} (\text{in Å}) = \frac{12400}{V (\text{in Volts})} = \frac{12400}{20000} = \frac{124}{200} = 0.62 \text{ Å} \] Comparing our result with the options:
- (A) 0.248 Å
- (B) 0.41 Å
- (C) 0.099 nm
- (D) 0.062 nm
Our calculated value matches option (D).
Step 4: Final Answer:
The minimum wavelength of the X-rays produced is 0.062 nm. Therefore, option (D) is correct.