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?

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")`