# H atom spectrum

1. Oct 30, 2005

### Pengwuino

Theres the question presented to me.

Now in all my wonderful glory, I enter "8" and boom, I am wrong. Just to be sure... i do 6, 7, 8, 9, and 10. None of them work. Then i start thinking... ok it can transfer from n=9 to n=8 to n=3 to n=1... but then that brings up an incredible number of combonations.

Am I right here with 8 or have i lost all comprehension of the H atom?

2. Oct 31, 2005

### Staff: Mentor

You're right, there's a lot of combinations of "hops" the electron can take in going from n = 9 down to n = 1. But the number of distinct, individual "hops" is actually fairly small. You should be able to enumerate them quickly.

3. Oct 31, 2005

### Pengwuino

So shouldn't it be roughly a 8+7+6+5+4+3+2+1 deal?

4. Oct 31, 2005

### daniel_i_l

If I correctly understand the question, you can get the number of wavelengths by:
$$n_{wavelengths} = \frac{n_{level} (n_{level} -1)}{2}$$