To determine the minimum speed at which the cesium plate must move downward so that the vertically upward component of the velocity of each emitted photoelectron becomes nonpositive, we need to consider the photoelectric effect and the energy dynamics involved. Let's break this down step by step.
Understanding the Photoelectric Effect
The photoelectric effect occurs when light of sufficient energy strikes a material and causes the emission of electrons. The energy of the incoming photons can be calculated using the formula:
Where:
- E is the energy of the photon in electron volts (eV).
- h is Planck's constant (approximately 4.1357 x 10-15 eV·s).
- c is the speed of light (approximately 3 x 108 m/s).
- λ is the wavelength of the incoming radiation.
Calculating Photon Energy
Given that the wavelength of the radiation is 250 nm (which is 250 x 10-9 m), we can calculate the energy of the photons:
- E = (4.1357 x 10-15 eV·s) * (3 x 108 m/s) / (250 x 10-9 m)
- E ≈ 4.97 eV
This energy is significantly higher than the work function of cesium, which is 1.9 eV. Thus, photons with a wavelength of 250 nm can indeed cause the emission of electrons from the cesium plate.
Energy Considerations for Photoelectrons
When a photon strikes the cesium plate and ejects an electron, the kinetic energy (KE) of the emitted electron can be expressed as:
Where φ is the work function (1.9 eV). Substituting the values, we find:
- KE = 4.97 eV - 1.9 eV = 3.07 eV
Relating Kinetic Energy to Velocity
The kinetic energy of an electron can also be expressed in terms of its mass and velocity:
Where m is the mass of the electron (approximately 9.11 x 10-31 kg). Setting the two expressions for kinetic energy equal gives us:
To convert eV to joules (1 eV = 1.6 x 10-19 J), we have:
- 3.07 eV = 3.07 x 1.6 x 10-19 J ≈ 4.91 x 10-19 J
Now, substituting the values into the kinetic energy equation:
- (1/2)(9.11 x 10-31 kg)v2 = 4.91 x 10-19 J
Solving for Velocity
Rearranging the equation to solve for v:
- v2 = (2 * 4.91 x 10-19 J) / (9.11 x 10-31 kg)
- v2 ≈ 1.08 x 1012 m2/s2
- v ≈ √(1.08 x 1012) ≈ 1.04 x 106 m/s
Considering the Downward Motion
For the upward component of the velocity of the emitted electrons to be nonpositive, the downward speed of the plate (v) must be at least equal to the speed of the emitted electrons. Therefore, the minimum value of v should be:
This means that the cesium plate must move downward at a speed of at least approximately 1.04 million meters per second to ensure that the upward velocity of the emitted photoelectrons does not exceed this downward speed, effectively making their vertical component nonpositive.
