- #1

John Greger

- 34

- 1

## Homework Statement

In He-like Carbon, C V, the excitation energy of 1s3s ##^1S## is 2851180 ##cm^-1## and for 1s4s ##^1S## it is 2988246 ##cm^-1##. Estimate the ionization energy. Compare with the value in the NIST database!

## Homework Equations

##E_{io}= T + E_{excitation}, ## ##T = ##\frac{R}{(n- \delta)^2} ## where delta is the quantum defect.

## The Attempt at a Solution

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I started with setting up the equations $$E_{io} = E_{exc}(3s) + \frac{R}{(3- \delta)^2}$$

$$E_{io} = E_{exc}(4s) + \frac{R}{(4-\delta)^2}$$

I subtracted the two equations so I could get rid of ##E_{io}## and just solve for delta, assuming that delta is the same for 3s and 4s.

Numerically in my pocket calculator I got that delta = ca 2.199.

I now plugged that into the equations for T, hence got a value for ##E_{io}##.

But that was wrong..

Apparently the quantum defect and ionization energy was very wrong.

I also assumed that the ionization energy was same fro 3s and 4s.

The answer is " With ##\delta## = 0.03 3s gives 3162180 cm-1 and 4s 3162300 cm-1"

My question is, what did I do wrong and how should I alternatively set up equations to solve for delta? If the ionization energy is not the same for 3s and 4s I can think of any other way.