Question

A monochromatic light beam with energy 204eV is used to excite the Be+3 (H –like atom with Z=4) from its ground state to some excited state m. Neglect the finite nuclear mass correction.
a) Find the quantum number m of this excited state.
b) Following this transition the excited Be+3 atom emits a spectral line at 30.42nm as a result of the transition from this excited state m to some other lower state n. Find the quantum number of this new state.
c) Calculate the orbital radius of the electron in the state n.

Answer 1

(a) For BeIII with Z=4, use the Rydberg equation for the energy levels. They will be 4^2 = 16 times more widely separated than for hydrogen.

E(m) = 13.58 eV*16/n^2 = 217.3 /m^2 eV

For a transition from the ground state (m=1),

204 eV = 217.3/1 - 217.3/m^2
217.3/m^2 = 16.4

m = 4 (probably) There may be some accuracy issues with the value of the Rydberg that I used.

(b) Convert the wavelength to photon energy in eV. Then use the Rydberg equation to compute n.

E = h*c/(wavelength) = ____ eV

E = 204 eV - R/n^2 = 204 - 217.3/n^2
Solve for n

(c) Bohr orbit radius = ao * n^2/Z
where ao is the hydrogen ground state Bohr radius