Question:
A photoelectric effect experiment is conducted with two different metal plates, A and B, using photons of two different frequencies, f1 and f2. The stopping potential for each frequency is measured and the results are given in the table below:
| Metal Plate | Frequency (f) | Stopping Potential (V) |
|---|---|---|
| A | f1 | 0.75 V |
| A | f2 | 1.25 V |
| B | f1 | 1.50 V |
| B | f2 | 3.00 V |
a) Calculate the difference in energy (ΔE) between the photons of frequency f1 and f2.
b) Determine the work function (Φ) and the threshold frequency (f0) for each metal plate.
Answer:
a) The difference in energy (ΔE) between two frequencies can be calculated using the equation:
ΔE = hf2 - hf1,
where h is the Planck's constant. We need to convert the frequencies to the corresponding wavelengths using the equation:
c = λf,
where c is the speed of light. The Planck's constant can be found in AP Physics 1 constants sheet as h = 6.63 × 10^-34 J·s.
Let's begin by calculating the wavelengths:
For frequency f1:
λ1 = c / f1,
For frequency f2:
λ2 = c / f2.
Substituting the values for speed of light (c = 3.00 × 10^8 m/s), frequency f1, and frequency f2 into the above equations, we can find the wavelengths λ1 and λ2.
Next, we can calculate the energy difference ΔE:
ΔE = (hc / λ2) - (hc / λ1),
Substituting the values of Planck's constant (h), speed of light (c), wavelength λ1, and wavelength λ2 into the above equation, we can now find the energy difference ΔE.
b) Work function (Φ) can be calculated using the equation:
Φ = eV,
where e is the elementary charge and V is the stopping potential.
Threshold frequency (f0) can be determined using the equation:
f0 = c / λ0,
where λ0 is the wavelength at which the stopping potential becomes zero.
Let's now calculate the work function and threshold frequency for each metal plate:
For metal plate A:
Using the stopping potential for f1 (0.75 V), we can calculate the work function (ΦA) by substituting the value of elementary charge (e) into the equation.
To find the threshold frequency (f0A), we need to determine the wavelength (λ0A) corresponding to the stopping potential of 0 V. We can do this by linear interpolation using the given stopping potentials for f1 and f2.
For metal plate B:
Using the stopping potential for f1 (1.50 V), we can calculate the work function (ΦB) by substituting the value of elementary charge (e) into the equation.
To find the threshold frequency (f0B), we need to determine the wavelength (λ0B) corresponding to the stopping potential of 0 V. We can do this by linear interpolation using the given stopping potentials for f1 and f2.
Let's calculate the values now.
Answer:
a)
Speed of light (c) = 3.00 × 10^8 m/s
Planck's constant (h) = 6.63 × 10^-34 J·s
Wavelength λ1 (for frequency f1) = c / f1
Substituting the values, we get:
λ1 = 3.00 × 10^8 m/s / (f1) = 3.00 × 10^8 m/s / (insert value of f1 here)
Wavelength λ2 (for frequency f2) = c / f2
Substituting the values, we get:
λ2 = 3.00 × 10^8 m/s / (f2) = 3.00 × 10^8 / (insert value of f2 here)
Energy difference ΔE = (hc / λ2) - (hc / λ1)
Substituting the values, we get:
ΔE = (6.63 × 10^-34 J·s * 3.00 × 10^8 m/s / λ2) - (6.63 × 10^-34 J·s * 3.00 × 10^8 m/s / λ1)
ΔE = (insert calculation here) J
b)
Work function (ΦA) = eV1 (for metal plate A, using f1)
Substituting the values, we get:
ΦA = (1.60 × 10^-19 C) * (insert value of V1 here) V = (insert calculation here) J
Threshold frequency (f0A) = c / λ0A (using the stopping potential of 0 V)
Wavelength λ0A can be determined by linear interpolation using the stopping potentials for f1 and f2:
λ0A = λ1 + (V0 * (λ2 - λ1) / (V2 - V1))
Substituting the values, we get:
λ0A = λ1 + (0 * (λ2 - λ1) / (V2 - V1)) = λ1 (as V0 = 0 V)
f0A = c / λ0A = c / (insert value of λ0A here) = (insert calculation here) Hz
For metal plate B:
Work function (ΦB) = eV1 (for metal plate B, using f1)
Substituting the values, we get:
ΦB = (1.60 × 10^-19 C) * (insert value of V1 here) V = (insert calculation here) J
Threshold frequency (f0B) = c / λ0B (using the stopping potential of 0 V)
Wavelength λ0B can be determined by linear interpolation using the stopping potentials for f1 and f2:
λ0B = λ1 + (V0 * (λ2 - λ1) / (V2 - V1))
Substituting the values, we get:
λ0B = λ1 + (0 * (λ2 - λ1) / (V2 - V1)) = λ1 (as V0 = 0 V)
f0B = c / λ0B = c / (insert value of λ0B here) = (insert calculation here) Hz
Therefore, the answer to part b is:
Work function for metal plate A (ΦA) = (insert calculated value here) J
Threshold frequency for metal plate A (f0A) = (insert calculated value here) Hz
Work function for metal plate B (ΦB) = (insert calculated value here) J
Threshold frequency for metal plate B (f0B) = (insert calculated value here) Hz