# 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

The first line Lyman series is between **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

The stationary hydrogen atom in its ground state has an *ionization energy* of *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")#