To solve this problem, we need to apply the concepts of the photoelectric effect and the de Broglie wavelength. The photoelectric effect describes how electrons are emitted from a metal surface when it is exposed to light (photons) of sufficient energy. The energy of the incoming photons is crucial in determining the kinetic energy of the emitted electrons. Let's break down the information given and find the work functions of metals A and B, as well as the wavelengths.
Understanding the Photoelectric Effect
According to the photoelectric effect, the energy of the incoming photon (E) is related to the work function (Φ) of the metal and the kinetic energy (KE) of the emitted electron by the equation:
E = Φ + KE
Where:
- E is the energy of the photon (in electron volts, eV).
- Φ is the work function of the metal (also in eV).
- KE is the kinetic energy of the emitted electron (in eV).
Given Data
From the problem, we have the following information:
- For metal A: Photon energy (x) = 4.25 eV, Kinetic energy of ejected electrons = $, Wavelength = #
- For metal B: Photon energy (y) = 4.70 eV, Kinetic energy of ejected electrons = $, Wavelength = #'
- It is given that @ = $ - 1.50 eV.
- Also, #' = 2#, meaning the wavelength of metal B is twice that of metal A.
Finding Work Functions
Let's denote the work function of metal A as Φ_A and that of metal B as Φ_B. We can set up the equations based on the energy of the photons and the kinetic energy of the emitted electrons.
For Metal A
Using the equation for metal A:
4.25 eV = Φ_A + $
For Metal B
Using the equation for metal B:
4.70 eV = Φ_B + ($ - 1.50 eV)
Substituting the expression for @:
4.70 eV = Φ_B + $ - 1.50 eV
Rearranging gives:
Φ_B = 4.70 eV - $ + 1.50 eV
Φ_B = 6.20 eV - $
Relating Work Functions
Now we have two equations:
- Φ_A = 4.25 eV - $
- Φ_B = 6.20 eV - $
To find the work functions, we can express Φ_B in terms of Φ_A:
Φ_B = Φ_A + 1.95 eV
Finding Wavelengths
Next, we need to find the wavelengths # and #'. The wavelength of a photon can be calculated using the equation:
λ = h / p
Where:
- h is Planck's constant (approximately 4.135667696 x 10^-15 eV·s).
- p is the momentum of the photon, which can be expressed as p = E/c (where c is the speed of light).
Since we know the energies of the photons, we can find the wavelengths:
For metal A:
# = h / (E/c) = hc / E
# = (4.135667696 x 10^-15 eV·s * 3 x 10^8 m/s) / 4.25 eV
For metal B, since #' = 2#, we can find #' using the same formula:
#' = 2 * #
Final Calculations
Now, substituting the values into the equations will allow us to calculate the work functions and wavelengths. You can use a calculator to find the exact numerical values for Φ_A, Φ_B, #, and #'.
In summary, by applying the principles of the photoelectric effect and the relationships between energy, work function, and wavelength, we can derive the necessary values for metals A and B. This approach not only helps in solving the problem but also reinforces the fundamental concepts of quantum mechanics and wave-particle duality.
