What would the minimum work function for a metal have to be for visible light ( 380 380 – 750 750 nm) to eject photoelectrons?

Hi everyone. In this practice problem, we're being asked to determine the minimum value of a work function. We will have an ultraviolet light with a wavelength range of nanometer to 400 nanometer incident on a lithium surface. We're being asked to determine the minimum value of the work function for the lithium surface. In order for it to be able to eject electrons, the options given R A 1.24 E V B 3.10 E V C 4.13 E V and lastly D 3.1 E V. So in order for us to be able to solve this problem, we will have to actually look at the threshold frequency. So threshold frequency or F T H will have to be used in order for us to be able to solve this problem. So we want to recall that the uh formula for actually calculating the work function or will be equals to H multiplied by F. And in this case, you want to employ the F which with FT H which is going to be the threshold frequency five is going to be the work function H is going to be the planks constant. So H is going to just be 4.136 times 10 to the power of negative 15 E V seconds. And after the H is the threshold frequency at the same time F T H will be equals to C divided by lambda T H where C is just the speed of light which is three times to the power of eight m per second. And lambda T H is just the threshold wavelength, we can substitute our F DH with this formula into our work function formula. So that we get uh Phi to be equals to H multiplied by after H with uh can be substituted with C divided by lambda DH. So Phi will be equals to H multiplied by C divided by lambda T H. So from this equation, we can conclude that the work function will actually be minimum when the wavelength is maximum because it is inversely correlated. So the larger the lambda is the smaller the work function is which is what we are interested to find, which is the minimum value of the work function. So lambda T H for the given range of 100 nanometer to nanometer will actually be equals to 400 nanometer, which is what we're going to be using for this formula. So then I can then be calculated by a multiply H which is 4.136 times 10 to the power of negative 15 E V seconds with C which is three times 10 to the power of eight m per second. And then dividing that with the lambda T H which is nanometer. But we have to convert that into meter by multiplying 400 with 10 to the power of negative nine m. And then calculating all of this, we will get our five value in terms of E V which will come out to be 3.10 E V. So 3.10 E V will be our minimum value for the work function for the lidum surface to eject electrons. And that will correspond to option B in our answer choices. So option B will be the answer to this particular practice problem. So that'll be it for this video. If you guys still have any sort of confusion, please make sure to check out our videos on similar topics available on our website and that'll be it for this one. Thank you.

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Problem 8

Textbook Question

What would the minimum work function for a metal have to be for visible light ( $380$ – $750$ nm) to eject photoelectrons?

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Understand the problem: The work function (Φ) is the minimum energy required to eject an electron from the surface of a metal. For visible light to eject photoelectrons, the energy of the photons in the visible spectrum (380–750 nm) must be equal to or greater than the work function. Use the relationship between photon energy and wavelength.

Recall the formula for the energy of a photon: $ E = \frac{hc}{\lambda} $, where $ h $ is Planck's constant ($ 6.626 \times 10^{-34} \, \text{J·s} $), $ c $ is the speed of light ($ 3.00 \times 10^8 \; \text{m/s} $), and $ \lambda $ is the wavelength of the photon in meters.

Determine the maximum photon energy in the visible spectrum: Use the shortest wavelength ($ \lambda = 380 \; \text{nm} $) to calculate the maximum energy. Convert $ \lambda $ to meters by dividing by $ 10^9 $. Substitute $ h $, $ c $, and $ \lambda $ into the formula $ E = \frac{hc}{\lambda} $.

Interpret the result: The maximum photon energy corresponds to the minimum work function (Φ) required to eject photoelectrons. This is because the photon energy must be at least equal to the work function for the photoelectric effect to occur.

Conclude: The minimum work function for the metal is equal to the maximum photon energy calculated in the previous step. Ensure the units are consistent (e.g., convert the result to electronvolts if needed, using $ 1 \; \text{eV} = 1.602 \times 10^{-19} \; \text{J} $).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Work Function

The work function is the minimum energy required to remove an electron from the surface of a material. It is a critical parameter in photoelectric effect experiments, as it determines the threshold frequency of light needed to eject electrons. The work function is typically expressed in electron volts (eV) and varies among different materials.

Photoelectric Effect

The photoelectric effect is the phenomenon where electrons are emitted from a material when it absorbs light of sufficient energy. According to Einstein's photoelectric equation, the energy of the incoming photons must be greater than or equal to the work function of the material for electron ejection to occur. This effect demonstrates the particle-like behavior of light and is foundational in quantum physics.

Photon Energy and Wavelength

The energy of a photon is inversely related to its wavelength, described by the equation E = hc/λ, where E is energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. For visible light, which ranges from 380 to 750 nm, the corresponding photon energies can be calculated, and these energies must meet or exceed the work function for photoelectrons to be emitted from the metal.

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What would the minimum work function for a metal have to be for visible light (380380–750750 nm) to eject photoelectrons?

Key Concepts

Work Function

Photoelectric Effect

Photon Energy and Wavelength

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What would the minimum work function for a metal have to be for visible light ( $380$ – $750$ nm) to eject photoelectrons?