Answer :
Sure, let's address each part of the question step-by-step:
### Part (a)
We need to determine the final de Broglie wavelength of an electron that accelerates from rest through a potential difference of 500 V.
1. Kinetic Energy Calculation: When the electron accelerates through a potential difference [tex]\( V \)[/tex], it gains kinetic energy (KE). The kinetic energy gained by the electron is given by:
[tex]\[ \text{KE} = eV \][/tex]
Here:
- [tex]\( e \)[/tex] is the charge of the electron, which is approximately [tex]\( 1.602 \times 10^{-19} \)[/tex] coulombs.
- [tex]\( V \)[/tex] is the potential difference, which is 500 volts.
Therefore, the kinetic energy of the electron is:
[tex]\[ \text{KE} = 1.602 \times 10^{-19} \, \text{C} \times 500 \, \text{V} = 8.01 \times 10^{-17} \, \text{J} \][/tex]
2. Velocity Calculation: The kinetic energy of the electron is also given by the classical equation:
[tex]\[ \text{KE} = \frac{1}{2} m v^2 \][/tex]
where:
- [tex]\( m \)[/tex] is the mass of the electron, which is approximately [tex]\( 9.109 \times 10^{-31} \)[/tex] kg.
- [tex]\( v \)[/tex] is the velocity of the electron.
Solving for [tex]\( v \)[/tex]:
[tex]\[ v = \sqrt{\frac{2 \times \text{KE}}{m}} = \sqrt{\frac{2 \times 8.01 \times 10^{-17} \, \text{J}}{9.109 \times 10^{-31} \, \text{kg}}} = 13261599.400632674 \, \frac{\text{m}}{\text{s}} \][/tex]
3. de Broglie Wavelength Calculation: The de Broglie wavelength [tex]\( \lambda \)[/tex] of a particle is given by:
[tex]\[ \lambda = \frac{h}{mv} \][/tex]
where:
- [tex]\( h \)[/tex] is Planck's constant, approximately [tex]\( 6.626 \times 10^{-34} \)[/tex] Js.
- [tex]\( m \)[/tex] is the mass of the electron as before.
- [tex]\( v \)[/tex] is the velocity we just calculated.
Therefore, the de Broglie wavelength [tex]\( \lambda \)[/tex]:
[tex]\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js}}{9.109 \times 10^{-31} \, \text{kg} \times 13261599.400632674 \, \frac{\text{m}}{\text{s}}} = 5.4851034724464483 \times 10^{-11} \, \text{m} \][/tex]
Thus, the electron's final de Broglie wavelength after accelerating through a potential difference of 500 V is [tex]\( 5.485 \times 10^{-11} \)[/tex] meters.
### Part (b)
Provide examples of experiments that reveal different behaviors of light and particles:
1. The Particle Nature of Light: The Photoelectric Effect is a classic experiment that demonstrates the particle nature of light. In this experiment, light shined onto a metal surface ejects electrons from the surface, but only if the light has a frequency above a certain threshold. This phenomenon can be explained by considering light as consisting of particles called photons, each with energy proportional to its frequency (as described by Einstein).
2. The Wave Nature of Particles: The Electron Diffraction Experiment demonstrates the wave-like nature of particles. When a beam of electrons is directed at a crystal, it creates a diffraction pattern similar to that produced by waves passing through a series of slits. This experiment provides direct evidence that particles such as electrons exhibit wave properties, confirming the de Broglie hypothesis.
In summary:
- (1) The particle nature of light: The Photoelectric Effect.
- (2) The wave nature of particles: Electron Diffraction Experiment.
### Part (a)
We need to determine the final de Broglie wavelength of an electron that accelerates from rest through a potential difference of 500 V.
1. Kinetic Energy Calculation: When the electron accelerates through a potential difference [tex]\( V \)[/tex], it gains kinetic energy (KE). The kinetic energy gained by the electron is given by:
[tex]\[ \text{KE} = eV \][/tex]
Here:
- [tex]\( e \)[/tex] is the charge of the electron, which is approximately [tex]\( 1.602 \times 10^{-19} \)[/tex] coulombs.
- [tex]\( V \)[/tex] is the potential difference, which is 500 volts.
Therefore, the kinetic energy of the electron is:
[tex]\[ \text{KE} = 1.602 \times 10^{-19} \, \text{C} \times 500 \, \text{V} = 8.01 \times 10^{-17} \, \text{J} \][/tex]
2. Velocity Calculation: The kinetic energy of the electron is also given by the classical equation:
[tex]\[ \text{KE} = \frac{1}{2} m v^2 \][/tex]
where:
- [tex]\( m \)[/tex] is the mass of the electron, which is approximately [tex]\( 9.109 \times 10^{-31} \)[/tex] kg.
- [tex]\( v \)[/tex] is the velocity of the electron.
Solving for [tex]\( v \)[/tex]:
[tex]\[ v = \sqrt{\frac{2 \times \text{KE}}{m}} = \sqrt{\frac{2 \times 8.01 \times 10^{-17} \, \text{J}}{9.109 \times 10^{-31} \, \text{kg}}} = 13261599.400632674 \, \frac{\text{m}}{\text{s}} \][/tex]
3. de Broglie Wavelength Calculation: The de Broglie wavelength [tex]\( \lambda \)[/tex] of a particle is given by:
[tex]\[ \lambda = \frac{h}{mv} \][/tex]
where:
- [tex]\( h \)[/tex] is Planck's constant, approximately [tex]\( 6.626 \times 10^{-34} \)[/tex] Js.
- [tex]\( m \)[/tex] is the mass of the electron as before.
- [tex]\( v \)[/tex] is the velocity we just calculated.
Therefore, the de Broglie wavelength [tex]\( \lambda \)[/tex]:
[tex]\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js}}{9.109 \times 10^{-31} \, \text{kg} \times 13261599.400632674 \, \frac{\text{m}}{\text{s}}} = 5.4851034724464483 \times 10^{-11} \, \text{m} \][/tex]
Thus, the electron's final de Broglie wavelength after accelerating through a potential difference of 500 V is [tex]\( 5.485 \times 10^{-11} \)[/tex] meters.
### Part (b)
Provide examples of experiments that reveal different behaviors of light and particles:
1. The Particle Nature of Light: The Photoelectric Effect is a classic experiment that demonstrates the particle nature of light. In this experiment, light shined onto a metal surface ejects electrons from the surface, but only if the light has a frequency above a certain threshold. This phenomenon can be explained by considering light as consisting of particles called photons, each with energy proportional to its frequency (as described by Einstein).
2. The Wave Nature of Particles: The Electron Diffraction Experiment demonstrates the wave-like nature of particles. When a beam of electrons is directed at a crystal, it creates a diffraction pattern similar to that produced by waves passing through a series of slits. This experiment provides direct evidence that particles such as electrons exhibit wave properties, confirming the de Broglie hypothesis.
In summary:
- (1) The particle nature of light: The Photoelectric Effect.
- (2) The wave nature of particles: Electron Diffraction Experiment.
