Physics Problem: X-ray Tube Calculations
Question:
X-rays are produced at an accelerating potential of 50 kV.
- 3.1 Find the shortest wavelength of the radiation.
- 3.2 Determine the corresponding frequency.
The X-ray tube has a potential difference of 40 kV and a current of 30 mA.
- 4.1 Find the number of electrons striking the target per second.
- 4.2 Determine the speed at which they strike.
- 4.3 Find the minimum wavelength of the X-ray produced.
Answer:
Given Constants:
- Planck’s constant (h) = 6.626 × 10⁻³⁴ J·s
- Speed of light (c) = 3.00 × 10⁸ m/s
- Charge of electron (e) = 1.602 × 10⁻¹⁹ C
- Electron mass (m) = 9.109 × 10⁻³¹ kg
3.1 Shortest Wavelength at 50 kV
Using the equation:
λmin = (hc) / (eV)
Plug in the values:
λmin = (6.626 × 10⁻³⁴ × 3.00 × 10⁸) / (1.602 × 10⁻¹⁹ × 5.0 × 10⁴) ≈ 2.48 × 10⁻¹¹ m = 0.0248 nm
3.2 Corresponding Frequency
Using:
f = c / λ
f = (3.00 × 10⁸) / (2.48 × 10⁻¹¹) ≈ 1.21 × 10¹⁹ Hz
4.1 Electrons per Second at 30 mA
Use:
n = I / e = 0.030 / (1.602 × 10⁻¹⁹) ≈ 1.87 × 10¹⁷ electrons/second
4.2 Speed of Electrons at 40 kV
Use energy conservation:
½ mv² = eV ⇒ v = √(2eV / m)
v = √[2 × 1.602 × 10⁻¹⁹ × 4.0 × 10⁴ / 9.109 × 10⁻³¹] ≈ 1.19 × 10⁸ m/s
4.3 Minimum Wavelength at 40 kV
λmin = (hc) / (eV) = (6.626 × 10⁻³⁴ × 3.00 × 10⁸) / (1.602 × 10⁻¹⁹ × 4.0 × 10⁴) ≈ 0.0311 nm
Final Answers Summary:
- 3.1 Minimum Wavelength (50 kV): 0.0248 nm
- 3.2 Frequency: 1.21 × 10¹⁹ Hz
- 4.1 Electrons per second: 1.87 × 10¹⁷ electrons/s
- 4.2 Speed of electrons: 1.19 × 10⁸ m/s
- 4.3 Minimum Wavelength (40 kV): 0.0311 nm