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### Topic: a problem on photoelectric effect  (Read 4712 times)

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#### Sona

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##### a problem on photoelectric effect
« on: April 12, 2017, 10:30:09 AM »
Sodium has an ionisation energy of 5.12eV. Calculate the energy of photoelectrons when light of wavelength 200nm shines on it. (work function of Na 2.3eV).

I have calculated the energy of radiation using the formula E=hc/λ and then converted it to eV unit. It is giving an value of 6.2. That means photoelectron energy should be (6.2-2.3), that is 3.9 eV.What is the use of ionisation energy here? Is it required to calculate the energy of photoelectron?

#### Enthalpy

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##### Re: a problem on photoelectric effect
« Reply #1 on: April 12, 2017, 01:52:00 PM »
A shortcut is: 1240nm in vacuum for 1eV.

In which state is your sodium? For a gas, atoms are separated, and the ionization energy applies. For a solid, the minimum energy is called the work function.

In the case of a solid, only a maximum photoelectron energy can be computed, because the electron can originate from an energy level deeper than the Fermi level.

There could be some subtleties with solids, for instance if the electrons of proper energy have non-zero momentum. They might need the help of a phonon then, and the process gets less efficient, meaning that the material needs more thickness to absorb light of a given energy. Or more generally, electrons have some energy-momentum relation in the metal, an other one in vacuum, photons have still an other one, and conserving the energy and the momentum in the interaction needs some angle conditions or the contribution of a phonon. Typical books exercises neglect this.

#### Sona

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##### Re: a problem on photoelectric effect
« Reply #2 on: April 13, 2017, 08:38:00 AM »
so if here Na is in gaseous state then the KE will be the difference between IE and energy of photon?

#### Enthalpy

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##### Re: a problem on photoelectric effect
« Reply #3 on: April 14, 2017, 03:24:55 PM »
That's how I understand it - with the proper signs.

As an extra subtlety, you might want to account the kinetic energy of the ion too, but it's small. Neglecting he photon's momentum, zero momentum sum for the electron plus the ion tells that the ion's energy is like 1800*23 times smaller, but then relativistic correction would matter.

#### Zeeshan Manzoor

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##### Re: a problem on photoelectric effect
« Reply #4 on: May 28, 2017, 02:38:04 PM »
but how can energy required to remove an electron from solid Na be lesser(you called work function=2.3eV) than the energy required to remove an electron from gaseous atom(I.E = 5.12eV)

#### Enthalpy

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##### Re: a problem on photoelectric effect
« Reply #5 on: May 31, 2017, 06:46:54 AM »
The figures I found confirm it: 2.28eV for the solid (it would depend on the crystal orientation) and 5.14eV for the gas
http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/photoelec.html
http://physics.nist.gov/PhysRefData/Handbook/Tables/sodiumtable1.htm

To hold the atoms together in a solid, one expects that the electrons get a lower energy, so that condensation produces energy. So the work function should exceed the ionization energy, seemingly. Is that the question?

My answer is then that the work function is the smallest possible energy that extracts an electron, that is, remove an electron that has the highest possible energy. Though, electrons in a metal have a wide range of energies spanning many eV. So while the maximum electron energy in metallic sodium can be 2.28eV below vacuum, the mean energy of shared electrons can be more than 5.14eV below vacuum, to make the solid energetically favourable over the gas. It only needs an energy range of populated states that is wide enough.

Then, one may ask why the mean electron energy in the solid could be below the energy in individual atoms, despite a usual understanding of orbitals interaction of two atoms is that it creates two molecular orbitals, one coupling and one anti-coupling, about equidistant energetically to the atomic orbitals. My elements of answer:
• Two electrons can populate one state, for instance both on a coupling molecular orbital. Same argument for the metallic bond as for the covalent one.
• "Energetically equidistant to the atomic orbitals" holds for loosely coupled atoms only.
• Metallic solids form one huge molecule of many more than 2 atoms and extending in three dimensions. This lowers the electrons' kinetic energy further, making the metal more favourable than a diatomic molecule.