A stationary #He^+# ion emitted a photon corresponding to a first line of the Lyman series. The photon liberated a photoelectron from a stationary H atom in the ground state. What is the velocity of the photoelectron?

1 Answer
Truong-Son N.
Sep 7, 2017

The first line Lyman series is between #n = 1# and #n = 2#. We know that #"He"^(+)# is isoelectronic with #"H"# atom, and thus, we can utilize the Rydberg equation for hydrogen-like atoms:

#DeltaE = -Z^2 cdot R_H (1/n_f^2 - 1/n_i^2)#

where:

  • #DeltaE# is the energy of the relaxed electron and thus of the emitted photon.
  • #Z# is the atomic number.
  • #R_H# is the Hydrogen ionization energy in the appropriate units (say, #"13.61 eV"#).
  • #n_k# is the principal quantum number for the #k#th electron state.

The energy that the emitted photon has due to the electronic relaxation is given by:

#DeltaE = -(2^2) cdot ("13.61 eV")(1/2^2 - 1/1^2)#

#=# #"40.83 eV"#

and would be pertaining to the #ul(2p -> 1s)# relaxation (due to the selection rules requiring the total orbital angular momentum to change as #DeltaL = pm1#).

The stationary hydrogen atom in its ground state has an ionization energy of #"13.61 eV"#, so its photoelectron would have a kinetic energy of:

#"40.83 eV" - "13.61 eV" = "27.22 eV"#

or #4.361 xx 10^(-18) "J"#, using the conversion factor #(1.602 xx 10^(-19) "J")/("1 eV")#.
(Why did I convert to #"J"#?)

As all electrons have mass, they also have velocities pertaining to the usual kinetic energy equation:

#K = 1/2 m_ev_e^2#

And thus, the forward velocity is given by:

#color(blue)(v_e) = sqrt((2K)/m_e)#

#= sqrt((2 cdot 4.361 xx 10^(-18) cancel"kg" cdot "m"^2"/s"^2)/(9.109 xx 10^(-31) cancel"kg")#

#= 3.094 xx 10^6 "m/s"#

#=# #ulcolor(blue)(3.094 xx 10^8 "cm/s")#

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