Photons striking a metal target

In summary, the metals sodium, aluminium and silver will not emit photoelectrons when irradiated by Balmer-B photons from a hydrogen discharge lamp.
  • #1
Andrusko
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Homework Statement



Sodium, aluminium and silver have work functions of 2.46, 4.08 and 4.73 eV respectively. Which, if any of these metals will emit photoelectrons when irradiated by Balmer-B photons from a hydrogen discharge lamp?


Homework Equations



[tex] Energy = 13.6(\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} [/tex]

The Attempt at a Solution



Using the above formula I have calculated that Balmer-B photons from a hydrogen discharge lamp have an energy of 2.55 eV.

My answer is none though, because I was under the impression that a photon has to give up all or none of its energy in a collision like this. So wouldn't the work function have to exactly match the photon energy?

That doesn't really sound right, but I'm sure that there is a condition about photons giving up all or none of their energy in this photoelectric stuff. Can someone please clarify?
 
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  • #2
Andrusko said:

Homework Statement



Sodium, aluminium and silver have work functions of 2.46, 4.08 and 4.73 eV respectively. Which, if any of these metals will emit photoelectrons when irradiated by Balmer-B photons from a hydrogen discharge lamp?


Homework Equations



[tex] Energy = 13.6(\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} [/tex]

The Attempt at a Solution



Using the above formula I have calculated that Balmer-B photons from a hydrogen discharge lamp have an energy of 2.55 eV.

My answer is none though, because I was under the impression that a photon has to give up all or none of its energy in a collision like this. So wouldn't the work function have to exactly match the photon energy?

That doesn't really sound right, but I'm sure that there is a condition about photons giving up all or none of their energy in this photoelectric stuff. Can someone please clarify?
The photon must indeed 'give up all or none' of it's energy to the electrons, but this doesn't mean that the photon energy needs to be equal to the work function. The work function is the energy required for the electron to become 'unbound' from the surface of the metal, any additional energy imparted on the electron by a photon contributes to the kinetic energy of the electron.

Does that make sense?
 
  • #3
Ah, of course! I see now. Thankyou very much!
 

FAQ: Photons striking a metal target

What is the process of photons striking a metal target?

Photons striking a metal target is also known as the photoelectric effect. This involves the interaction of photons, which are particles of light, with the electrons in a metal. When a photon strikes the metal surface, it transfers its energy to an electron, causing it to be ejected from the metal.

How does the number of photons affect the number of electrons ejected?

The number of photons striking a metal target is directly proportional to the number of electrons ejected. This means that the more photons that hit the target, the more electrons will be ejected. This relationship is known as the photoelectric effect equation: E = hf, where E is the energy of the photon, h is Planck's constant, and f is the frequency of the light.

What is the work function of a metal and how does it affect the photoelectric effect?

The work function of a metal is the minimum amount of energy required to remove an electron from the surface of the metal. If the energy of the incoming photon is greater than the work function, then the electron will be ejected. If the energy of the photon is less than the work function, then no electrons will be ejected.

Can the photoelectric effect be observed with any type of light?

No, the photoelectric effect can only be observed with light that has a high enough frequency to eject electrons from the metal surface. This is why the photoelectric effect was originally observed with ultraviolet light, which has a higher frequency than visible light. However, with the advent of more powerful light sources, the photoelectric effect can now be observed with visible light as well.

What is the significance of the photoelectric effect?

The photoelectric effect has significant implications in various fields of science and technology. It helped to establish the concept of quantization of energy and was a key factor in the development of quantum mechanics. It also has practical applications such as in photovoltaic cells, which convert light energy into electricity, and in photocells, which are used in light sensors and detectors.

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