Finding the wavelength from the work function in photoemission

[EmmaLemming]

Homework Statement

a) Write down the formula that relates the maximum electron energy, Emax. to the
frequency of the incident light in the photoelectric effect.

b) Calculate the maximum wavelength of light for which photoemission occurs for light
incident on a metal whose workfunction is 2.30 eV.


2. The attempt at a solution

a) E = hf - σ

where E is the maximum energy, h is Planck's constant and σ is the work function.

b) E = (hc/λ) - σ

where λ is the wavelength and c is the speed of light in a vacuum

∴ λ = hc/(E+σ)

However I'm not given the energy so how am I to calculate λ?

Is there another way to calculate E?
I looked in my textbook but found nothing relevant :(

 

[collinsmark]

Homework Statement

a) Write down the formula that relates the maximum electron energy, Emax. to the
frequency of the incident light in the photoelectric effect.

b) Calculate the maximum wavelength of light for which photoemission occurs for light
incident on a metal whose workfunction is 2.30 eV.2. The attempt at a solution

a) E = hf - σ

where E is the maximum energy, h is Planck's constant and σ is the work function.

b) E = (hc/λ) - σ

where λ is the wavelength and c is the speed of light in a vacuum

∴ λ = hc/(E+σ)

However I'm not given the energy so how am I to calculate λ?

The problem statement asked for "maximum wavelength". Don't forget, maximum wavelength corresponds to minimum frequency; thus minimum energy.

What happens if the wavelength is so large that hc/λ = σ ? What's the wavelength that would cause photoemission to occur, but without any leftover energy?

 

[EmmaLemming]

Okay, so I assume that minimum energy is 0 and then write,

hc/σ = λ

I get, λ = 5.4 x 10^-7 m

That seems reasonable , right?

I don't know the answer to your last question.

 

[collinsmark]

Okay, so I assume that minimum energy is 0 and then write,

hc/σ = λ

I get, λ = 5.4 x 10^-7 m

That seems reasonable , right?

'Looks right to me.

I don't know the answer to your last question.

What I mean is that if the light's wavelength is too big, there is not enough energy to overcome the work function and photo-emission doesn't take place at all. As you decrease the wavelength, the energy increases. Decrease it enough and photo-emission begins to occur. Decrease the wavelength even more and the leftover photon energy can end up becoming electron energy.

So the maximum wavelength is the wavelength where the maximum electron energy is zero.

 

[asteriatic]

I would first clarify the information provided in the problem. It states that the work function of the metal is 2.30 eV, which is a unit of energy. Therefore, we can use this value to calculate the maximum energy of the emitted electrons.

The formula E = hf - σ is used to calculate the maximum energy of the emitted electrons, where h is Planck's constant and f is the frequency of the incident light. We can rearrange this formula to solve for the frequency, as follows:

f = (E + σ) / h

Now, we can use this frequency to calculate the wavelength of the incident light using the formula λ = c/f, where c is the speed of light in a vacuum. This gives us the maximum wavelength of light for which photoemission occurs.

However, if we are not given the energy of the emitted electrons, we can still calculate the maximum wavelength using the work function and the speed of light. The work function, σ, is defined as the minimum amount of energy required to remove an electron from the surface of the metal. Therefore, the energy of the incident light must be greater than or equal to the work function in order for photoemission to occur.

We can use this information to set up an inequality:

E ≥ σ

This means that the energy of the incident light must be greater than or equal to the work function. Rearranging this inequality gives us:

E + σ ≥ 0

Now, we can use this value in the formula λ = hc/(E+σ), as follows:

λ = hc/(E+σ) ≥ hc/0

Since division by zero is undefined, this means that the maximum wavelength of light for which photoemission occurs is infinite. This makes sense, as any wavelength of light with enough energy to overcome the work function will result in photoemission. Therefore, the maximum wavelength is not a specific value, but rather a range of values with no upper limit.

In conclusion, the formula E = hf - σ can be used to calculate the maximum energy of the emitted electrons, which can then be used to calculate the maximum wavelength of light for photoemission. However, if the energy is not given, we can still determine that the maximum wavelength is infinite using the work function and the speed of light.