Atomic Structure: Impact of 663nm Photon on Excited Hydrogen Atom

Click For Summary

Homework Help Overview

The discussion revolves around the interaction of a 663nm photon with a hydrogen atom in its fourth excited state (n=5). Participants are exploring whether this photon can ionize the atom and what the resulting kinetic energy of the electron would be after ionization.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are using the Rydberg formula to determine the conditions for ionization and are questioning the energy associated with the given wavelength. They are also discussing the implications of different values for n1 and n2 in the context of ionization.

Discussion Status

There is an ongoing exploration of the energy of photons at different wavelengths, particularly comparing 663nm and the threshold wavelength for ionization (2278nm). Some participants are clarifying the relationship between wavelength and energy, while others are seeking to understand the kinetic energy of the electron post-ionization.

Contextual Notes

Participants are working within the constraints of the problem, specifically regarding the excited state of the hydrogen atom and the implications of photon energy relative to ionization thresholds.

apunisheriii
If a photon of wavelength 663nm is incident on a hydrogen atom in its fourth excited state(n=5),then
A.the atom doesn't absorb the photon
B.the atom absorbs the photon but is not ionized
C.the atom absorbs the photon and is ionized,with the electron having zero kinetic energy after ionization
D.the atom absorbs the photon and is ionized,with the electron having kinetic energy 1.33eV after ionization.

the answer is D

but i couldn't approach tat answer when I am using 1/lambda=R(1/n1-1/n22)
after sub in the values into this equation
whole thing seems to be wrong...as 1/25-1/25=0

how can i know tat the photon with wavelength 663nm is able to ionize the atom??
and how to find the electron's kinetic energy after ionization?
 
Physics news on Phys.org
1/λ = R (1/n12 - 1/n22)

Start with, what wavelength would ionize the atom? (Use the above equation, thinking carefully about what n1 and n2 would be).

How much energy does a photon of that wavelength have?

Then, does a 663 nm photon have at least that much energy?
 
let n2=0 and n1=25
then i get the wavelength as 2278nm
this is the wavelength for ionization of tat atom
rite??
so it means tat 663nm couldn't ionize the atom??
 
apunisheriii said:
let n2=0 and n1=25

Actually, we let 1/n22=0 and 1/n12=25
(n2=∞ and n1=5)

then i get the wavelength as 2278nm
this is the wavelength for ionization of tat atom
rite??

Yes. Or more correctly, that's the threshhold wavelength for ionization of the atom. Any photons with more energy could also ionize the atom.

so it means tat 663nm couldn't ionize the atom??

You should figure out the energy in photons of 2278nm and 663nm in order to understand what is going on. Or, realize that photons have more energy if the wavelength is ___________. (longer or shorter?)

p.s. I'm logging off soon, good luck!
 
oo!
i got it !
thnx!
but how bout the kinetic energy?
 
Last edited by a moderator:

Similar threads

Replies
4
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 23 ·
Replies
23
Views
5K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K