In which option is the highest velocity of electron?

a) #"H"#, #n=2#
b) #"He"^(+)#, #n=6#
c) #"Li"^(2+)#, #n=3#
d) #"Be"^(3+)#, #n=5#

1 Answer
Truong-Son N.
Nov 24, 2017

Do you not expect it to be the one with the highest ionization energy? That one would leave its "orbit" with the highest velocity, as #K prop v^2#.

Well, you should realize that all of these atoms have one electron. Thus, they are hydrogenic atoms, with energies

#E_n = -"13.61 eV" cdot Z^2/n^2#

where #Z# is the atomic number and #n# is the principal quantum number.

So, each of these electronic energies are...

#E_2("H") = -"13.61 eV" cdot 1^2/2^2 = -"3.403 eV"#

#E_6("He"^+) = -"13.61 eV" cdot 2^2/6^2 = -"1.512 eV"#

#E_3("Li"^(2+)) = -"13.61 eV" cdot 3^2/3^2 = -"13.61 eV"#

#E_5("Be"^(3+)) = -"13.61 eV" cdot 4^2/5^2 = -"8.710 eV"#

And using the fact that #1.602 xx 10^(-19) "J" = "1 eV"#, these energies in #"J"# are:

#ul(E_2("H")) = -3.403 cancel"eV" xx (1.602 xx 10^(-19) "J")/(cancel"1 eV")#

#= ul(5.452 xx 10^(-19) "J")#

#ul(E_6("He"^+)) = -1.512 cancel"eV" xx (1.602 xx 10^(-19) "J")/(cancel"1 eV")#

#= ul(2.422 xx 10^(-19) "J")#

#ul(E_3("Li"^(2+))) = -13.61 cancel"eV" xx (1.602 xx 10^(-19) "J")/(cancel"1 eV")#

#= ul(2.180 xx 10^(-18) "J")#

#ul(E_5("Be"^(3+))) = -8.710 cancel"eV" xx (1.602 xx 10^(-19) "J")/(cancel"1 eV")#

#= ul(1.395 xx 10^(-18) "J")#

While it is physically inaccurate to specify each electron's velocity (it violates the Heisenberg Uncertainty Principle!), one would supposedly calculate it by using the energy input as the output energy of the electron as it "travels" towards #n = oo# to begin to escape the atom.

#|E_n| = K = 1/2mv^2#

#=> |v| = sqrt((2K)/m_e)#

As a result, each of these excited electrons "travel" at the following speeds in the Bohr model of the atom towards #n = oo# (remembering that the premise of this problem is completely false!):

#ul(v_2("H")) = sqrt((2 cdot 5.452 xx 10^(-19) "J")/(9.109 xx 10^(-31) "kg"))#

#= ul(1.094 xx 10^6 "m/s")#

#ul(v_6("He"^(+))) = sqrt((2 cdot 2.422 xx 10^(-19) "J")/(9.109 xx 10^(-31) "kg"))#

#= ul(7.292 xx 10^(5) "m/s")#

#color(blue)(ul(v_3("Li"^(2+)))) = sqrt((2 cdot 2.180 xx 10^(-18) "J")/(9.109 xx 10^(-31) "kg"))#

#= color(blue)ul(2.188 xx 10^(6) "m/s")#

#ul(v_5("Be"^(3+))) = sqrt((2 cdot 1.395 xx 10^(-18) "J")/(9.109 xx 10^(-31) "kg"))#

#= ul(1.750 xx 10^(6) "m/s")#

As expected, the electron which had to receive the most ionization energy would move the fastest.

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