• Support PF! Buy your school textbooks, materials and every day products Here!

Photoelectric effect and Rydberg's formula

  • Thread starter iuchem16
  • Start date
  • #1
4
0
A peak in the energy spectrum is seen at 134.2 keV when using a germanium detector. In coincidence with this, an x-ray corresponding to the 2s to 1s electronic transition in germanium is detected. Find the energy (in keV) of the gamma ray that ejected the electron.

Use Rydberg's formula for spectral lines of hydrogenic atoms:

1/lambda = R Z2 ( ( 1/n12) - ( 1/n22) )

where R, the Rydberg constant, is such that Rhc = 13.61 eV
(c is the speed of light, h is Planck's constant).




not really sure where to begin, do I solve for the wavelength and use that in the energy equation E=hc/lambda??
 

Answers and Replies

  • #2
mgb_phys
Science Advisor
Homework Helper
7,774
12
You might want to substitute the above equation into the equation for energy that you mentioned. Especiay since you are given R in terms of Rhc
 
  • #3
4
0
nevermind..i got it...thanks!
 

Related Threads on Photoelectric effect and Rydberg's formula

  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
9
Views
5K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
1K
Replies
6
Views
4K
Replies
5
Views
2K
Replies
0
Views
3K
  • Last Post
Replies
0
Views
785
Top