The Student Room Group
Reply 1
millano
The ionisation energy of hydrogen is 13.6 ev

What is the speed of the slowest electron that can ionise a hyrdoegen atom when it collides with it?

I dont know what forumla to use - i think its something that we havent covered.

Thanks in advance


(1/2)(m_e)(v^2)=2.18E-18

lvl=+sqrt((4.36E-18)/(m_e))
lvl=4.78E12 m(s^-1)

RPFeynman.
Reply 2
so whats the answer, sorry i dont understand that

apparently the answer it gives is: 2.19(10^6) ms-1
Reply 3
millano
The ionisation energy of hydrogen is 13.6 ev

What is the speed of the slowest electron that can ionise a hyrdoegen atom when it collides with it?

I dont know what forumla to use - i think its something that we havent covered.

Thanks in advance


(1/2)(m_e)(v^2)=2.18E-18

lvl=+sqrt((4.36E-18)/(m_e))

lvl=2.19E6 m(s^-1)

RPFeynman.
Reply 4
l0l i dont understnad what you wrote there one bit (i cant really read it l0l)
Reply 5
millano
l0l i dont understnad what you wrote there one bit (i cant really read it l0l)


to ionise the hydrogen atom you need to give it 13.6eV of energy. You ccan do that by using the kinetic energy carried by the electron you fire at it.

13.6eV = (1/2)mv^2

you know the m, the mass of an electron, so just rearrange to find v, the velocity of the electron.

13.6eV = 13.6 * 1.6(10^-19) J = .5 * 9.1(10^-31) * v^2

v = 2.19(10^6)ms-1
Reply 6
oh yh (1/2) mv^2

thx
If you had a 13.6eV electron come in and knock out an electron from the H atom, giving it 13.6eV, you'd end up with the initial incoming electron having no energy, and would therefore be captured by the resultant proton surely?

Wouldn't the minimal energy be at least double the ionisation energy, so that both electrons have enough energy to climb out of the potential well they are in when they are that close to the proton? If the incoming electron had say 20eV, there would not be enough energy to get both electrons "to infinity", an thus ionise the atom.
Reply 8
AlphaNumeric
If you had a 13.6eV electron come in and knock out an electron from the H atom, giving it 13.6eV, you'd end up with the initial incoming electron having no energy, and would therefore be captured by the resultant proton surely?

Wouldn't the minimal energy be at least double the ionisation energy, so that both electrons have enough energy to climb out of the potential well they are in when they are that close to the proton? If the incoming electron had say 20eV, there would not be enough energy to get both electrons "to infinity", an thus ionise the atom.


theorectically, no this would not be the case. Think about it. If an electron is free it has 0 energy. We model the situation as an electron in a well, 13.6eV deep. We bring along an electron, it has +13.6eV. We do a bit of jiggery pokery and end up with two electrons, both with 0 energy. Both are technically free of the atom. In reality of course, to do this you have to take the electrons to infinity, which just aint gonna happen.

I think the confusion is in how the energy is transferred to the orbitting electron. Our moving electron starts at infinity, it starts to move in towards the atom. Ignoring any interactions with the orbiting electron, our moving electron starts to GAIN kinetic energy (it's falling in to the well....so it must be getting faster, the energy has to go somewhere). When it finally hits the atom it has 27.2eV of energy, just enough to get itself and the orbitting electron back out. (we assume the electron radiates no energy as it approaches the atom as well, which is of course false)
Ah good point, it'll gain the required energy to escape as it falls, so it's just 13.6eV. Silly me :p: