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.

Answers

drwls
(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

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