Number of Wavelengths from Excited H Atom (n=9)

  • Thread starter Thread starter Pengwuino
  • Start date Start date
  • Tags Tags
    Atom Spectrum
Click For Summary
SUMMARY

The discussion centers on calculating the number of distinct wavelengths emitted by a hydrogen atom's electron transitioning from the n=9 energy level. The correct formula for determining the number of wavelengths is given by n_{wavelengths} = n_{level} (n_{level} - 1) / 2, which results in 36 distinct wavelengths. Participants initially miscalculated by considering individual transitions without recognizing the combinatorial nature of the problem. The correct approach involves understanding the transitions between energy levels and applying the formula accurately.

PREREQUISITES
  • Understanding of quantum mechanics and energy levels in atoms
  • Familiarity with the hydrogen atom and its electron transitions
  • Knowledge of combinatorial mathematics
  • Basic grasp of spectral lines and wavelengths
NEXT STEPS
  • Study the derivation of the formula for calculating wavelengths in hydrogen transitions
  • Explore quantum mechanics principles related to electron transitions
  • Learn about spectral lines and their significance in atomic physics
  • Investigate combinatorial methods in physics problems
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in atomic structure and quantum mechanics, particularly those focusing on hydrogen atom behavior and spectral analysis.

Pengwuino
Gold Member
Messages
5,109
Reaction score
20
Assume that a hydrogen atom's electron has been excited to the n = 9 level. How many wavelengths of light can be emitted as this excited atom loses energy?

there's 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?
 
Physics news on Phys.org
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.
 
So shouldn't it be roughly a 8+7+6+5+4+3+2+1 deal?
 
If I correctly understand the question, you can get the number of wavelengths by:
n_{wavelengths} = \frac{n_{level} (n_{level} -1)}{2}
 

Similar threads

Replies
16
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 0 ·
Replies
0
Views
621
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K