Photoelectrons ejecting from Cs metal

[utkarshakash]

Homework Statement

A small piece of cesium metal (W=1.9eV) is kept at a distance of 20cm from a large metal plate having a charge density of 1.0*10^-9 C m^-2 on the surface facing the cesium piece. A monochromatic light of wavelength 400nm is incident on the cesium piece. Find the minimum and the maximum kinetic energy of the photoelectrons reaching the large metal plate.

Homework Equations

The Attempt at a Solution

I can find the electric field due to metal plate and thus the corresponding potential difference. I can also find the stopping potential. But what about kinetic energies?

 

[STEMucator]

I believe to find $E_{k_{max}}$ you should use:

$E_k = \frac{hc}{\lambda} - W$ - Don't forget to convert your work function into Joules and nm to meters.

If you knew the cutoff voltage, there would be another way to find the max kinetic energy as well by using the charge density.

Also, to find $E_{k_{min}}$, you should think about what causes the minimum kinetic energy. Think about the threshold frequency and how it relates to the minimum kinetic energy. The light must meet this minimum frequency in order to give enough energy to the photoelectrons, so that they can be ejected from the cesium metal to the large metal plate.

 

[rude man]

Since there is only one radiating frequency there will be one value of k.e. corresponding to the difference between hf and the given work function W. That would have to be the max. k.e. also.

To determine the min. k.e. it seems one would have to know the distribution of the work function for Cesium among its atoms, which I could not find. This may instead have to do with the energy levels of the electrons in each atom. The min. W would be associated with the outermost orbital electrons, which would = 1.9 eV, and the max. W with the innermost orbital electrons, for which W = outermost W + difference in energy betw. outermost & innermost electrons. Need expert physics help here ...

 

[utkarshakash]

I believe to find $E_{k_{max}}$ you should use:

$E_k = \frac{hc}{\lambda} - W$ - Don't forget to convert your work function into Joules and nm to meters.

If you knew the cutoff voltage, there would be another way to find the max kinetic energy as well by using the charge density.

Also, to find $E_{k_{min}}$, you should think about what causes the minimum kinetic energy. Think about the threshold frequency and how it relates to the minimum kinetic energy. The light must meet this minimum frequency in order to give enough energy to the photoelectrons, so that they can be ejected from the cesium metal to the large metal plate.

I did my calculation and got KE(max) = 1.19eV which is not correct.

 

[rude man]

I did my calculation and got KE(max) = 1.19eV which is not correct.

You did not include the effect of the electric field. You just computed the photon energy and subtracted the work function.

BTW are you sure the charge on the plate is positive? It's usually negative in experiments.

The serious question of minimum k.e. remains ...

 

[utkarshakash]

You did not include the effect of the electric field. You just computed the photon energy and subtracted the work function.

BTW are you sure the charge on the plate is positive? It's usually negative in experiments.

The serious question of minimum k.e. remains ...

But the question mentions that electric field is to be neglected (sorry for not giving that info earlier).

 

[ehild]

hc/λ-W=KE(max), the maximum kinetic energy. That is the KE of the electrons which escaped the metal from the top of the potential well. Other electrons can have less KE than that. The minimum KE of the electrons leaving the metal is zero. They are accelerated by the electric field on the surface charge of the opposite metal plate . That surface charge density is 1.0*10^-9 C m^-2 positive (there would be a minus sign in front otherwise). What is the electric field between the metal plate and the Caesium piece?

ehild

 

[rude man]

But the question mentions that electric field is to be neglected (sorry for not giving that info earlier).

Why should the electric field be neglected if it was given in great detail?

 

[rude man]

hc/λ-W=KE(max), the maximum kinetic energy. That is the KE of the electrons which escaped the metal from the top of the potential well. Other electrons can have less KE than that. The minimum KE of the electrons leaving the metal is zero. They are accelerated by the electric field on the surface charge of the opposite metal plate . That surface charge density is 1.0*10^-9 C m^-2 positive (there would be a minus sign in front otherwise). What is the electric field between the metal plate and the Caesium piece?

ehild

You have not explained how electrons can have zero emitted k.e. Why zero exactly?

A photon at 400 nm imparts 3.1 eV to an electron, which exceeds the work function by 1.2 eV. How can an electron be kicked to the surface with zero k.e. or indeed with any k.e. < 1.2 eV? Must have something to do with the quantized electron energy levels: S, P etc. And if it does, the lowest released electron's k.e. is not likely to be zero since the orbital electron levels are quantized. Would be quite a coincidence.

Please explain.

 

[ehild]

You have not explained how electrons can have zero emitted k.e. Why zero exactly?

A photon at 400 nm imparts 3.1 eV to an electron, which exceeds the work function by 1.2 eV. How can an electron be kicked to the surface with zero k.e. or indeed with any k.e. < 1.2 eV? Must have something to do with the quantized electron energy levels: S, P etc. And if it does, the lowest released electron's k.e. is not likely to be zero since the orbital electron levels are quantized. Would be quite a coincidence.

Please explain.

The energy levels in a crystal are arranged in bands. The energy levels in the band, although quantized, are very close to each other: the atomic levels split when the atoms interact. In case of two interacting atoms, they split into two, in case of N atoms in a crystal, they split into N sub-levels. The outermost electrons occupy levels in the valence band, those of the metals occupy levels in the conduction band. There are electrons with energy near the bottom of the band and there are others near the Fermi level.

Have you ever asked yourself why the maximum kinetic energy appears in the equation hf=W+KE(max)?

KE(max) is the upper bound for the KE of the electrons kicked out of the metal, so there must be ones with lower KE. What can be the lower bound of the KE? Anyway, it can not be negative.

ehild

 

[rude man]

Have you ever asked yourself why the maximum kinetic energy appears in the equation hf=W+KE(max)?

KE(max) is the upper bound for the KE of the electrons kicked out of the metal, so there must be ones with lower KE. What can be the lower bound of the KE? Anyway, it can not be negative.

ehild

Of course it cannot be negative, but my question was, what is it?

But OK, what I have found is that

work function phi = E₀ - E_F

where E₀ is the energy required to kick an electron out of the lowest free electron state, and E_F is the Fermi level. Since E_F for Cesium = 1.6 eV and phi = 1.9 eV, this indicates that E₀ ~ 3.5 eV for Cs. There is a continuum of electron energies with values from - 3.5 eV to -1.9 eV so there is effectively a continuum of work functions also, from 1.9 eV to 3.5 eV. The work function given (1.9 eV) is the energy required to liberate an electron in the highest free electron state; all lower-state electrons require more than 1.9 eV. The lowest-state electrons require a light frequency f corresponding to hf = 3.5 eV. This frequency is above our light emitter frequency of c/λ, λ = 400 nm.

So, bottom line, the lowest emitted electron k.e. is indeed zero since the 400 nm radiation can liberate electrons only to the - 3.1 eV level.

So now the OP can compute the range of kinetic energies at the charged plate.