Question

The shortest wavelength, present in X-rays produced by an accelerating potential of 50kV, is:

• A. 25 Å
• B. 2.5 Å
• C. 0.25 Å
• D. 0.025 Å

Hint:

The formula \( \lambda_{\text{min}} (\text{Å}) \approx \frac{12400}{V (\text{volts})} \) is extremely useful and a major time-saver in exams. Memorize it. The constant 12400 comes from the product \(hc/e\) in units of eV·Å.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
When electrons are accelerated through a potential difference V and strike a target, they produce a continuous spectrum of X-rays (bremsstrahlung radiation). The shortest possible wavelength (\(\lambda_{\text{min}}\)) is produced when an electron loses all its kinetic energy in a single interaction, creating a single X-ray photon of maximum energy. This is described by the Duane-Hunt law.
Step 2: Key Formula or Approach:
The kinetic energy of an electron accelerated through a potential V is \( E = eV \).
The energy of a photon is \( E = hf = \frac{hc}{\lambda} \).
For the shortest wavelength, the entire kinetic energy of the electron is converted into the energy of one photon:
\[ eV = \frac{hc}{\lambda_{\text{min}}} \]
\[ \lambda_{\text{min}} = \frac{hc}{eV} \]
A very useful shortcut for this calculation is:
\[ \lambda_{\text{min}} (\text{in Å}) \approx \frac{12400}{V (\text{in volts})} \]
Step 3: Detailed Explanation:
Given the accelerating potential \( V = 50 \) kV \( = 50 \times 10^3 \) V \( = 50000 \) V.
Using the shortcut formula:
\[ \lambda_{\text{min}} (\text{Å}) = \frac{12400}{50000} \]
\[ \lambda_{\text{min}} (\text{Å}) = \frac{124}{500} = 0.248 \text{ Å} \]
Step 4: Final Answer:
The calculated value of 0.248 Å is closest to the option 0.25 Å.