- #1

- 15

- 0

Li+(g) ---> Li+2 + e-

Equation: frequency = (3.29x10^15s-1)Z^2(1/n^2i - 1/n^2f)

Ei + hv = Ef

My question: These equations can give me the energy of each energy level but how do I find the energy of required for ionization??

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter wee00x
- Start date

- #1

- 15

- 0

Li+(g) ---> Li+2 + e-

Equation: frequency = (3.29x10^15s-1)Z^2(1/n^2i - 1/n^2f)

Ei + hv = Ef

My question: These equations can give me the energy of each energy level but how do I find the energy of required for ionization??

- #2

Borek

Mentor

- 28,658

- 3,153

Ionization means energy is higher than the one needed for the electron to jump from n=1 to n=∞.

- #3

- 15

- 0

But then how would that be calculated?

- #4

Borek

Mentor

- 28,658

- 3,153

For very large n 1/n^{2} is for all practical purposes equal to zero.

- #5

- 12,121

- 160

That equation only works when there is one electron. If there are two or more, you pretty much have to measure (or be told) what the ionization energy is.Equation: frequency = (3.29x10^15s-1)Z^2(1/n^2i - 1/n^2f)

So, in what situation does the above equation apply to lithium?

Last edited:

Share: