Grade 12th passMechanics
Determine the frequency of revolution of the electron in 2nd Bohr’s orbit in hydrogenatom.
2 Answers
Arun
Approved Tutor Answer8 Years agoDear Aditya
Frequency of revolution = 6.55 * 10^15 z²/n³
Put z = 1 and n = 2..hence
Frequency = 8.187 * 10^14 sec -1
Khimraj
Approved Tutor Answer8 Years agoFrequency of electron revolving in orbit is given by
f = (me4/4n3h3
)(z2/n3)
Now n = 2 and z = 1
f = (6.55*1015)(12/23)
f = 8.187*1014 s-1
Hope it clears.
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