An electron of energy 10.8 ev is incident on an H atom then?

  1. the electron will come out with 10.8 ev energy
  2. the electron will completely absorbed
  3. 10.2 ev will be absorbed and it will come out with 0.6 ev energy
  4. none

1 Answer
Truong-Son N.
Feb 26, 2018

That is way more energy than necessary to make "H"^(-)(g)H−(g), so I suppose the electron will be absorbed... and then spit back out.

A hydrogen atom has an electron affinity of -"72.8 kJ/mol"−72.8 kJ/mol, or -"0.755 eV"−0.755 eV.

![http://2012books.lardbucket.org/](useruploads.socratic.org)

As a result, by accepting an electron,

"H"(g) + e^(-) -> "H"^(-)(g)H(g)+e−→H−(g)

the electron is stabilized by "0.755 eV"0.755 eV. So the electron will be left with

"10.80 eV" - "0.755 eV" = "10.05 eV"10.80 eV−0.755 eV=10.05 eV

to use. As the electron affinity of "H"(g)H(g) is -"0.755 eV"−0.755 eV, the ionization energy of "H"^(-)(g)H−(g) is "0.755 eV"0.755 eV, so the electron will then bounce back out with

"10.05 eV" - "0.755 eV" = "9.29 eV"10.05 eV−0.755 eV=9.29 eV

of energy. So... none of the above.

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