wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A 100 eV electron collides with a stationary helium ion (He+) in its ground state and excites to a higher level. After the collision, He+ ion emits two photons in succession with wavelengths 1085˚A and 304˚A. Find the principal quantum number of the excited state. Also, calculate the energy of the electron after the collision. Given h=6.63×1034J s.

Open in App
Solution

Energy of He+ in ground state =13.612×4=54.4
Total energy before collision =10054.4=45.6eV
From conservation of energy:
45.6=E+hc1085+hc304E=45.611.4240.78E=45.652.2E=6.613.6n2×4=6.6n2=8.2n2
Principal quantum number = 2

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Photoelectric Effect
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon