Q ] Determine the frequency[f] of revolution of an electron in the second orbit of helium atom ANS:- Data - Chemistry -

Question

Q ] Determine the frequency[f] of revolution of an electron in the
second orbit of helium atom
ANS:- Data :
z = 2, n= 2, nth =2nd , v = 2 18 x
106 ms-1 , find frequency(f)
Now,
n= v/ 2 pie r
n = 2 18 x 106 ms-1 / 2 x 3 14 x 0 529 x
10-10 m
n = [ 2 18 x 106 / 2 x 3 14 x 0 529 ] s-1
n = [ 2 18 x 1016 / 2 x 3 14 x 0 529 ]
s-1
n = 6 562 x 1015 [ f of atom in ground state of
hydrogen atom ]
Hence,
n = 6 562 x 1015 x 2
n = 13 124 x 105 (Hz)
[ is this correct method ? am i in right direction? ]

Avni · Accepted Answer

What is n in the solution? The solution is given below:

ν=cλλ=hmvSo, ν=cmvhc=velocity of lighth= Planck's constantm=mass of electronv=velocity of electron in nth orbit
Velocity of electron in 2nd orbit of He,
vn=2 18×106×Znm/sv2=2
18×106×22 m/s    =2
18×106 m/s
ν=cmvhν=3×108 ms-1×9
1×10-31 kg×2 18×106 ms-16
626×10-34 kgm2s-2 s  =8 98×1017 s-1