What is the Hydrogen Emission Spectrum for Transitions to the n = 1 Level?

Click For Summary
SUMMARY

The discussion focuses on the hydrogen emission spectrum, specifically the wavelengths emitted during electron transitions to the n = 1 energy level. The relevant equation used is 1/λ = Rh[1/(m^2)-1/(n^2)], with Rh = 1.09 x 10^7 m^-1. The transitions yield wavelengths primarily in the ultraviolet range, with specific calculations showing emissions at 91.6 nm and 4.86 nm for transitions from higher quantum levels down to n = 1. It is confirmed that quantum numbers m and n must be whole numbers due to the quantization of energy levels in hydrogen.

PREREQUISITES
  • Understanding of quantum mechanics and energy levels
  • Familiarity with the Rydberg formula for hydrogen
  • Knowledge of electromagnetic spectrum and wavelength ranges
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the Rydberg formula in detail, focusing on its applications to hydrogen and other elements
  • Learn about the electromagnetic spectrum, specifically the characteristics of ultraviolet light
  • Explore quantum mechanics principles, particularly quantization of energy levels
  • Investigate the implications of non-integral quantum numbers in quantum mechanics
USEFUL FOR

Students studying quantum mechanics, physics educators, and anyone interested in the spectral analysis of hydrogen and other atomic structures.

d.tran119
Messages
2
Reaction score
0

Homework Statement


What wavelengths emitted from a hydrogen gas discharge tube are associated with transitions from higher levels down to the n = 1 level?
[a] infrared
visible
[c] mixture of infrared and visible
[d] ultraviolet



Homework Equations


Equations:
1/λ = Rh[1/(m^2)-1/(n^2)] Rh = 1.09 x 10^7 m^-1.



The Attempt at a Solution


Can someone explain hydrogen gas electron transition to me? This stuff is a little over my head.

I obtained various wavelengths with arbitrary quantum numbers greater than 1. I took +infinity as a bound since higher quantum numbers reaches the series limit.
1/λ*(1 m/10^ 9 nm) = (1.097e-7 m^-1)[1/(1^2)-1/(infinity)]
λ (+infinity ,1)= 91.6 nm

1/λ*(1 m/10^ 9 nm) = (1.097e-7 m^-1)[1/(1^2)-1/(4^2)]
λ (4 ,1)= 4.86 nm

The level transitions yield a photon with wavelengths corresponding to UV light.

I’m confused though because taking bounds of the 1.01 type down to 1 allows the photon to be associated with different lights. Do I have to keep m and n whole when doing this problem? With m & n being non-integral values n can be taken arbitrarily closer and closer to 1 (with n>1) pushing the wavelength to positive infinity. Non-integral values possesses a different number of points in a dimensional space corresponding to different emissions.
 
Physics news on Phys.org
d.tran119 said:

Homework Statement


What wavelengths emitted from a hydrogen gas discharge tube are associated with transitions from higher levels down to the n = 1 level?
[a] infrared
visible
[c] mixture of infrared and visible
[d] ultraviolet

Homework Equations


Equations:
1/λ = Rh[1/(m^2)-1/(n^2)] Rh = 1.09 x 10^7 m^-1.

The Attempt at a Solution


Can someone explain hydrogen gas electron transition to me? This stuff is a little over my head.

I obtained various wavelengths with arbitrary quantum numbers greater than 1. I took +infinity as a bound since higher quantum numbers reaches the series limit.
1/λ*(1 m/10^ 9 nm) = (1.097e-7 m^-1)[1/(1^2)-1/(infinity)]
λ (+infinity ,1)= 91.6 nm

1/λ*(1 m/10^ 9 nm) = (1.097e-7 m^-1)[1/(1^2)-1/(4^2)]
λ (4 ,1)= 4.86 nm

The level transitions yield a photon with wavelengths corresponding to UV light.

I’m confused though because taking bounds of the 1.01 type down to 1 allows the photon to be associated with different lights. Do I have to keep m and n whole when doing this problem? With m & n being non-integral values n can be taken arbitrarily closer and closer to 1 (with n>1) pushing the wavelength to positive infinity. Non-integral values possesses a different number of points in a dimensional space corresponding to different emissions.


Yes, you have to keep m and n whole numbers. That's quantization. There are only discrete states for bound electrons.
 
Okay I understand this now. Thanks a lot!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K