posted by Sandhya .
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
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/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)
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.
Correction: 12.23 eV is less than the ioniztion energy. So, it is abound state problem as explained by Bobpursley
I don't follow your steps. It is asked about longest & shortest possible wavelengths.Can you please explain?Thanks