SUMMARY
The minimum wavelength of an electron accelerated through a potential difference of 30,000 V can be calculated using the equations E = qV and E = hf. The total energy of the electron is determined to be 4.8 x 10^-15 J. By manipulating the equation E = hf into E = hc/λ, the minimum wavelength can be derived, which occurs at maximum energy. This confirms that the minimum wavelength corresponds to the maximum energy of the electron as it accelerates.
PREREQUISITES
- Understanding of basic physics concepts such as energy, voltage, and wavelength
- Familiarity with the equations E = qV and E = hf
- Knowledge of Planck's constant (h) and the speed of light (c)
- Basic understanding of electron behavior in electric fields
NEXT STEPS
- Research the implications of electron acceleration in different potential differences
- Learn about the relationship between energy, frequency, and wavelength in quantum mechanics
- Explore the concept of wave-particle duality and its relevance to electrons
- Investigate applications of electron acceleration in technologies such as cathode ray tubes and particle accelerators
USEFUL FOR
Students studying physics, educators teaching quantum mechanics, and professionals in fields related to electronics and particle physics.