Atomic physics

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5)a. Electrons accelerated by a potential difference of 12.23 V pass through a gas of hydrogen atoms at room temperature. Calculate the wavelength of light emitted with the longest possible wavelength.
b. Calculate the wavelength of light emitted with the shortest possible wavelength

* physics - bobpursley, Sunday, April 13, 2008 at 12:53pm

Energy= 13.6*(1/n^2-1/1^2) which is the energy absorbed in a ground state to n state.
12.23>13.6*(1/n^2-1) or
12.23>13.6/n^2 - 13.6
-13.6/n^2 > -1.11
n^2< 12.34 or max n is 3

Energy= 13.6(1/9-1)=12.08 eV

Now from this, you can calculate max wavelength...

Energy= hf= hc/lambda compute lambda (remember to change eV to joules).

Energy= hf= hc/lambda

lambda=hc/13.2 * n^2)

* physics - Count Iblis, Sunday, April 13, 2008 at 1:06pm

The elctrons have an energy of 12.23 eV. The ionization energy of hydrogen is 13.6 eV. So, the electron beam will create ionized hydrogen as well as hydrogen in excited states.

The shortest wavelength photons are emitted when an 12.23 eV electron from the beam is captured by an ionized hydrogen atom (that atom would then be ionized by an earlier collision with an electron from the beam) and bound in the ground state. The energy of the photon is then 12.23 eV + 13.6 eV = 25.83 eV which corresponds to a wavelength of 48 nm.

There is no upper limit on the wavelength of the photon that can be emitted. When an electron from the beam collides with an hydrogen atom, the bound electron can be kicked out at any energy. These elctrons can be captured by protons at arbitrary high quantum number n. This means that the emitted photon can have arbitrary low energies.

Also, the electrons in the gas will interact with the hydrogen atoms and this will cause emission of photons at arbitrary low energies.

* physics: Correction - Count Iblis, Sunday, April 13, 2008 at 1:10pm

Correction: 12.23 eV is less than the ioniztion energy. So, it is abound state problem as explained by Bobpursley

* physics - Sandhya, Monday, April 14, 2008 at 7:41am

Hello Bobpursley

• Atomic physics - ,

Bob Pursley has shown that the highest value of principle quantum number n that can be excited from the n=1 state by a 12.23 eV electron is n=3. All of the room temperature hydrogen would be in the ground state. It would also be H2, and not H atoms, but the electron beam would dissociate some of the H2.

The longest wavelngth that would be emitted following an excitation to that level would be the 3-> 2 Balmer line, which is red and at 656.3 nm. Its wavelength can be calulated with the Rydberg formula. The shortest wevelength observable would correspond to the 3-> 1 Lyman series line in the very far "vacuum" ultraviolet, at 102.5 nm. You wold not be able to see this line using a spectrometer that operates in air, because air strongly absorbs the radiation.

This analysis neglects the possibility of multi-step electronic excitation, which can lead to higher quantum levels and both longer and shorter wavelength emission.

• Atomic physics - ,

Thanks drwls & Bob Pursley for your gerat help.

• Atomic physics - ,

Thanks drwls & Bob Pursley for your great help.