Which Energy Levels in Hydrogen Produce the Red Line?

Click For Summary
The prominent red line in the hydrogen spectrum corresponds to a wavelength of 6.60 x 10-7 m, which equates to an energy of approximately 3.0E-19 J. Calculations show that a transition from the next allowed energy level above the ground state to the ground state releases 5.40E-19 J, which does not match the energy of the red line. Additionally, a transition from a free electron to the next allowed energy level results in an energy release of 1.64E-18 J, also exceeding the energy of the red line. Therefore, it is concluded that the lowest energy level, identified as the ground state, cannot produce the prominent red line. The discussion emphasizes the importance of energy level transitions in understanding hydrogen's emission spectrum.
catkin
Messages
215
Reaction score
0

Homework Statement


This is from Advanced Physics by Adams and Allday, part 8 Modern Physics, Practice Exam Question 15.

  • The spectrum of atomic hydrogen contains a prominent red line having a wavelength of 6.60 x 10-7 m. Calculate the energy of a photon with this wavelength.
  • The ionisation energy of hydrogen is 2.18 x 10-18 J. The next allowed energy level above the ground state in hydrogen has an energy -5.40 x 10-19 J. Show by calculation that the lowest energy level cannot be involved in the production of the prominent red line in a.

Homework Equations


E = h f
v = f λ

The Attempt at a Solution


f = v / λ
E = h v / λ

= 6.63E-34 * 3.00E+8 / 6.60E-7
= 3.0E-19 J ct2sf (Book gives same answer. Calculated 3.013636364e-19)

A drop from the next allowed energy level above the ground state to the ground state would release 5.40E-19 J. This does not match energy calculated in a so this drop cannot be the one that produces the red line having a wavelength of 6.60E-7 m.

A free electron dropping to the next allowed energy level above the ground state would release 2.18E-18 - 5.40E-19 = 1.64E-18 J. This is more than the energy that produces the red line having a wavelength of 6.60E-7 m so I cannot, on the available data, show that some drop down to the next allowed energy level above the ground state does not produce the red line having a wavelength of 6.60E-7.

Hmm ... :confused:
 
Physics news on Phys.org
I believe by lowest energy level the question means the ground state, since this is the lowest allowed energy level. If this is the case, then you can show that the observed emission line cannot possibly be cause by a transition to the ground state.
 
Ah! Thank you Hootenanny :smile:

Then an answer to b is:

The ground state is the lowest energy level. When an excited electron drops from the next allowed energy level to the ground state the energy given off is 2.18E-18 - 5.40E-19 = 1.64E-18 J. This is more than the energy calculated in a so this drop cannot be the one that produces the red line in a. Drops from other allowed energy levels to ground state give off more energy so also cannot produce the red line in a. Thus the lowest energy level cannot be involved in the production of the prominent red line in a.
 
catkin said:
Ah! Thank you Hootenanny :smile:

Then an answer to b is:

The ground state is the lowest energy level. When an excited electron drops from the next allowed energy level to the ground state the energy given off is 2.18E-18 - 5.40E-19 = 1.64E-18 J. This is more than the energy calculated in a so this drop cannot be the one that produces the red line in a. Drops from other allowed energy levels to ground state give off more energy so also cannot produce the red line in a. Thus the lowest energy level cannot be involved in the production of the prominent red line in a.
Sounds spot on to me :approve:
 
Thanks :smile:
 
catkin said:
Thanks :smile:
A pleasure :smile:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
11K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
3
Views
3K
Replies
2
Views
2K
Replies
4
Views
3K
Replies
3
Views
2K