Hg vapor energy levels & Frank-Hertz experiment

[bentzy]

All who are familiar with this famous experiment, know that multiples of 4.9 Volts are used, which is associated with the difference between the ground state and one that belong to the next level. The question is it related to this transition (only ?), rather than also to any of its multiplet ? and even more, why not to higher levels transitions ?

 

[Parlyne]

Transitions to higher energy levels will occur; but, they will be extremely rare compared with transitions to the first excited state. This can be understood as follows. To get to the second excited state or higher it is necessary either that a single atom either collides with two electrons in such rapid succession that the atom does not have sufficient time to radiate down to the ground state, or that the atom collides with an electron that has enough energy to cause a transition directly from the ground state to the second excited state. This second case requires that the electron not collide with an atom between the point at which its energy reaches 4.9 eV and the point at which its energy reaches the energy difference between the ground and second excited state of Hg (I don't know this value off the top of my head). This will be very rare given how short the mean free path in the vapor is compared with the length needed for this extra energy acquisition.

 

[bentzy]

I'm afraid my point was misunderstood. First, your second phrase from the end of your reply, isn't clear. Second, when we increase gradually the voltage, the elctrons acquire higher kinetic energy, and the current drops right after 4.9, 9.8 etc. Now suppose we are at a voltage which corresponds to transition to a higher energy level, between 4.9 & 9.8. Is this transition and others alike, significantly less probable ? If we do the same with Hydrogen gas
in low pressure, the emission spectrum is composed of all allowed transitions, and the corresponding lines are strong !; this is also true when you do basic grating spectroscopy of a Hg vapor lamp - you see clearly enough 4 lines !

 

[Parlyne]

If we run electrons through the Frank-Hertz tube without the Hg gas, we'll find the electrons' kinetic energy is pretty much a linear function of how far they've traveled down the tube from the cathode. This means that the electrons must travel a certain distance to gain a given amount of kinetic energy. If the driving voltage is V, and the tube length is L, the electron's energy when it's at point x should be $E = e V \frac{x}{L}$.

If the energy difference between the first and second excited states of Hg is $E_{12}$, the electron will need to travel a distance $d = L \frac{E_{12}}{e V}$ down the tube after it already has enough energy to excite an atom to the first excited state to be able to excite one to the second excited state. However, if this electron collides with an atom before it travels this extra distance, it is likely to excite that atom to the first excited state and, thus, never have enough energy to excite and atom to the second excited state.

This means that, in order to cause a large enough number of transitions to the second excited state to detect, it must not be overwhelmingly likely that the electron encounters and atom before it travels a distance d. The typical distance an electron can travel in the gas before a collision is called the "mean free path." So, if the mean free path is much smaller than d, it should be rare that any atom gets excited to the second excited state by a single electron.

The Frank-Hertz tube is generally designed to have a rather short mean free path, unlike the low pressure tubes used for spectroscopy.

 

[sah-sah]

in frank hertz expriment with Hg vapor,in our digram I(current) in terms of accelerated potential, we have some sharp drops,that each drop corresponds with integral multiple of excitation energy, means V=n.(4.9ev).where n is integer.

how to explain it?

 

[bentzy]

My predecessor's reply isn't clear. Is it a question, or an emphasis of some sort ?

 

[sah-sah]

its a question.
can you reply me?

My predecessor's reply isn't clear. Is it a question, or an emphasis of some sort ?

 

[sah-sah]

in frank hertz expriment with Hg vapor,in our digram I(current) in terms of accelerated potential, we have some sharp drops,that each drop corresponds with integral multiple of excitation energy, means V=n.(4.9ev).where n is integer.

how to explain it?

If we run electrons through the Frank-Hertz tube without the Hg gas, we'll find the electrons' kinetic energy is pretty much a linear function of how far they've traveled down the tube from the cathode. This means that the electrons must travel a certain distance to gain a given amount of kinetic energy. If the driving voltage is V, and the tube length is L, the electron's energy when it's at point x should be $E = e V \frac{x}{L}$.

If the energy difference between the first and second excited states of Hg is $E_{12}$, the electron will need to travel a distance $d = L \frac{E_{12}}{e V}$ down the tube after it already has enough energy to excite an atom to the first excited state to be able to excite one to the second excited state. However, if this electron collides with an atom before it travels this extra distance, it is likely to excite that atom to the first excited state and, thus, never have enough energy to excite and atom to the second excited state.

This means that, in order to cause a large enough number of transitions to the second excited state to detect, it must not be overwhelmingly likely that the electron encounters and atom before it travels a distance d. The typical distance an electron can travel in the gas before a collision is called the "mean free path." So, if the mean free path is much smaller than d, it should be rare that any atom gets excited to the second excited state by a single electron.

The Frank-Hertz tube is generally designed to have a rather short mean free path, unlike the low pressure tubes used for spectroscopy.

 

[SpectraCat]

in frank hertz expriment with Hg vapor,in our digram I(current) in terms of accelerated potential, we have some sharp drops,that each drop corresponds with integral multiple of excitation energy, means V=n.(4.9ev).where n is integer.

how to explain it?

Have you tried looking in wikipedia? The basic answer to your question is there, but if you need further help, you can always come back and ask a more specific question.

 

[MartinV05]

I don't think you understood his question correctly. This has been troubling me too. Let's say an electron with 5.4 eV collides with an Hg atom and the energy needed to move an electron from the ground level to the second level is 5.4 eV. Why will the electron continue to move with speed reduced by 4.86 eV (the one that the atom of Hg absorbs) and not absorb it all to move the ground level electron to the second excited level.

 

[Parlyne]

I don't think you understood his question correctly. This has been troubling me too. Let's say an electron with 5.4 eV collides with an Hg atom and the energy needed to move an electron from the ground level to the second level is 5.4 eV. Why will the electron continue to move with speed reduced by 4.86 eV (the one that the atom of Hg absorbs) and not absorb it all to move the ground level electron to the second excited level.

If the electron has enough energy to excite an electron to the second excited state it can do so (although, the probability of such a transition is lower than 100%). The point of my post from almost 2 years ago is that Franck-Hertz tubes are designed such that collisions where the electron has enough energy for this to occur will be very rare.

 

[MartinV05]

Thank you for the reply even after 2 years of inactivity of this thread. But I'm sorry to say, I still don't get it. I don't see how the tube can prevent it from happening. And the line
"Franck-Hertz tubes are designed such that collisions where the electron has enough energy for this to occur will be very rare." is quite confusing because the experiment itself contains a grid which accelerates them to a point of 2*E, 3*E, n*E, and in one moment it WILL have the same amount of energy as the one needed to move it to the second level.

 

[Redbelly98]

... in one moment it WILL have the same amount of energy as the one needed to move it to the second level.

No, not if the electron collides inelastically with a mercury atom before getting to that energy. The electron loses 4.9 eV of kinetic energy when such a collision happens, that is why the electron doesn't (or very rarely) gets to a high enough energy to excite the next excited state.

In other words, before the electron ever gets to the required 10.2 eV energy, it loses 4.9 eV in colliding and exciting an atom to the 1st excited state.