1B.15 states that The velocity of an electron that is emitted from a metallic surface by a photon is 3.6 x 10^3 km/s.
(a) What is the wavelength of the ejected electron?
(b) No electrons are emitted from the surface of the metal until the frequency of the radiation reaches 2.50 x 10^16 Hz. How much energy is required to remove the electron from the metal surface?
(c) What is the wavelength of the radiation that caused photoejection of the electron?
(d) What kind of electromagnetic radiation was used?
I've solved for parts a and b but what is the best way to approach part c?
1B.15 Part C
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Re: 1B.15 Part C
Hi,
The wavelength of the radiation that caused the photoejection refers to the wavelength of the incoming photon.
After solving for part b, you should have found the work function. Then, from the given velocity, you would be able to find the kinetic energy of the ejected electron using Ek= 1/2 mv^2. After finding the energy for the ejected electron(Ek) and the energy needed to remove an electron(work function), you can use the
E(phtoton) - work function = Ek to solve for the energy of the incoming photon.
Once you find the energy of the photon, you can use to find the wavelength using E=hv.
The wavelength of the radiation that caused the photoejection refers to the wavelength of the incoming photon.
After solving for part b, you should have found the work function. Then, from the given velocity, you would be able to find the kinetic energy of the ejected electron using Ek= 1/2 mv^2. After finding the energy for the ejected electron(Ek) and the energy needed to remove an electron(work function), you can use the
E(phtoton) - work function = Ek to solve for the energy of the incoming photon.
Once you find the energy of the photon, you can use to find the wavelength using E=hv.
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- Posts: 100
- Joined: Fri Sep 24, 2021 5:04 am
Re: 1B.15 Part C
We know from part b the minimum energy needed to eject an electron (threshold energy or work function) and can use the work function to find the energy needed to eject an electron with a velocity of 3.6x10^3 km/s.
(1/2)mv^2 = (1/2)*(9.11x10^-31)*(3.6x10^6)^2 = 5.903x10^-18 J
hv - work function = (1/2)mv^2
hv = (1/2)mv^2 + work function
hv = 5.903x10^-18 J + 1.66x10-17 J = 2.25 x10^-17 J
Then, we can find the wavelength of the photon.
E = (ch)/λ
λ = (ch)/E
(1/2)mv^2 = (1/2)*(9.11x10^-31)*(3.6x10^6)^2 = 5.903x10^-18 J
hv - work function = (1/2)mv^2
hv = (1/2)mv^2 + work function
hv = 5.903x10^-18 J + 1.66x10-17 J = 2.25 x10^-17 J
Then, we can find the wavelength of the photon.
E = (ch)/λ
λ = (ch)/E
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