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

What are the frequency and wavelength of a photon during a transition from n= 5 state to n = 2 state in the He+ ion?

Open in App
Solution

Energy between two states = E = (13.6 eV) [1nf2-1ni2]
Energy of nth state = ​En = (-13.6 eV)/n2

So, E5 = (-13.6eV)/52 = -0.544 eV
E2 = (-13.6eV)/22= -3.4 eV

Energy during transition from n= 5 state to n = 2 state,
E = (13.6 eV) [1n22-1n52]
= E2-E5
=
-3.4 eV + 0.544 eV​
= -2.856 eV
​Therefore on going from level 5 to level 2, electron loses 2.856 eV of energy.


Converting this to joules gives E = 2.856 × 1.60 x 10-19 J/eV = 4.5696 x 10-19 J

Now, E = hcλ

λ=hcE = 6.63×10-34 × 3×1084.5696 ×10-19
​ = 4.3527×10-7 m

The frequency,
ν=cλ = 3×1084.3527×10-7 = 6.89 s-1


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Threshold Frequency
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon