Electrons are ejected from a metallic surface with speeds ranging up to 4.50x10^5 m/s when light with a wavelength of 615nm is used. (a) What is the work function of the metal? (b) What is the cutoff frequency for this surface?
Einstein's photoelectric equation:
ε =W + KE,
h•c/λ = W +m•v²/2,
W = h•c/λ - m•v²/2 =
= 6.63•10^-34•3•10^8/615•10^-9 - 9.1•10^-31•(4.5•10^5)/2 =
= 2.3•10^-19 J = 1.44 eV.
W = h•f,
f = W/h = 2.3•10^-19/6.63•10^-34 = 3.48•10^14 Hz.
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(a) Well, well, well, let's figure out the work function of this metal surface. So, we know that the energy of a photon is given by Planck's equation, E = hf, where h is the famous Planck's constant. Now, the energy of the ejected electron is given by E = KE + W, where KE is the kinetic energy of the ejected electron and W is our desired work function. Hold on tight, we're about to unleash some math!
We can express the energy of the photon in terms of its wavelength using the equation E = hc/λ, where c is the speed of light, and λ is the wavelength. This photon energy must be equal to the sum of the work function and the kinetic energy of the electron, so we have hc/λ = KE + W.
Since we're dealing with light and electrons here, we know that the kinetic energy of the ejected electron can be expressed as KE = (1/2)mv², where m is the mass of the electron and v is its velocity. We know the velocity, which is 4.50x10^5 m/s, but we don't need to bother with the mass of the electron since it cancels out in our calculations. Fancy, isn't it?
Alright, let's plug in the values and solve for the work function. hc/λ = (1/2)m(4.50x10^5)² + W. Given that λ is 615 nm, convert it to meters by dividing by 10^9: λ = 615x10^-9 m. Now, rearrange the formula to solve for W: W = hc/λ - (1/2)m(4.50x10^5)².
Crunch all those numbers, and you'll finally find the work function of the metal surface. Ta-da!
(b) Now that we've got the work function sorted, let's find the cutoff frequency! The cutoff frequency is the minimum frequency at which electrons can escape from the metal surface.
We can bring in Einstein's photoelectric equation here: E = hf = Φ + KE, where E is the energy of the photon, f is the frequency of the light, Φ is the work function, and KE is the kinetic energy of the ejected electron.
We know the energy of a photon can be expressed as E = hc/λ, where λ is the wavelength. Now, the cutoff frequency is the frequency at which the kinetic energy becomes zero, so we get 0 = Φ + KE.
Rearranging our equation, we have Φ = -KE. But since KE = (1/2)m(v_cutoff)², where v_cutoff is the cutoff velocity, we can rewrite Φ = -m(v_cutoff)²/2.
Now we can introduce the equation v_cutoff = fλ, where f is the cutoff frequency. Substituting this into our equation, we have Φ = -m(fλ)²/2.
Now, to find the cutoff frequency, let's solve for f: f = sqrt(2Φ/mλ²).
Plug in the value of the work function you calculated earlier, the mass of the electron, and the wavelength, and you'll have your cutoff frequency. Good luck with those number-crunching acrobatics!
To find the work function of the metal and the cutoff frequency for the surface, we can make use of the photoelectric effect equation:
1. Work Function (Φ):
The work function (Φ) can be determined using the equation:
E = hf = Φ + KE
where E is the energy of the incident photon, hf is the energy of the incident photon (h represents Planck's constant and f is the frequency), Φ is the work function, and KE is the kinetic energy of the emitted electron.
We know the speed of the emitted electrons (v = 4.50 × 10^5 m/s) and the wavelength of the incident light (λ = 615 nm). To find the energy of the incident photon, we can use the equation:
E = hc/λ
where h is Planck's constant and c is the speed of light.
Calculate the energy of the incident photon using the equation and convert it to joules:
E = (6.626 × 10^-34 J s)(3.00 × 10^8 m/s)/(615 × 10^-9 m)
E ≈ 3.23 × 10^-19 J
For electrons, the kinetic energy can be calculated using the equation:
KE = 1/2 mv^2
where m is the mass of an electron (9.11 × 10^-31 kg) and v is the speed of the emitted electrons.
Now, solve the equation for the work function (Φ):
Φ + KE = E
Φ = E - KE
Substitute the values and calculate:
Φ ≈ (3.23 × 10^-19 J) - (1/2)(9.11 × 10^-31 kg)(4.50 × 10^5 m/s)^2
Φ ≈ 2.77 × 10^-19 J
Therefore, the work function of the metal is approximately 2.77 × 10^-19 J.
2. Cutoff Frequency (f_cutoff):
The cutoff frequency (f_cutoff) is the minimum frequency of light required to eject an electron from the surface of a metal when no external potential is applied.
The cutoff frequency can be calculated using the equation:
f_cutoff = c/λ_cutoff
where c is the speed of light and λ_cutoff is the cutoff wavelength.
Convert the given wavelength from nanometers to meters:
λ_cutoff = 615 nm × (1 m/10^9 nm)
λ_cutoff ≈ 6.15 × 10^-7 m
Use the equation to calculate the cutoff frequency:
f_cutoff = (3.00 × 10^8 m/s)/(6.15 × 10^-7 m)
f_cutoff ≈ 4.88 × 10^14 Hz
Therefore, the cutoff frequency for this surface is approximately 4.88 × 10^14 Hz.