There's an easier way to do this but I don't have it yet so here is the long way.
1/wavelength = RZ^2(1/n^2 - 1/n^2)
Solve for wavelength.
R = 1.0973E7
Z = 4
The first n^2 in the formula is 1
The second n&2 in the formula is 0
So 1.0973E7*4^2((1-0) and that gives you 1/wavelength in meters. Plug wavelength into and solve for E in J/atom
Multiply that by 6.022E23 to convert to J/mol and convert to kJ/mol.
Then E = hc/wavelength
Calculate the fourth ionization energy of beryllium in J atom^-1 & kJ mol^-1 units
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