# PLEASE HELP!!!!!!!!! really easy... atomic physics calculate the angular frequency of an electron occupying the second bohr orbit of He+ ion? the answer is2.07x10^16 (units=second inverse)

2
by liz98

2015-05-17T04:32:57+05:30

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Let
v = velocity of an electron
m = mass of an electron = = 9.1  * 10^-31 kg
Z = atomic number = number of protons
k_e = Coulombs constant = 1 / (4 pi epsilon) = 9 * 10^9 units
e = charge on an electron = 1.6 * 10^-19 Coulombs
r = radius of revolution of  an electron around nucleus
n = principal quantum number of the electron = Bohr's orbit number  = 1,2, 3, 4 etc.
w = angular frequency of revolution of electron around nucleus
h = Planck' constant = 6.626 * 10^{-34}  units
h'  = h / 2 pi  =   1.054 * 10^{-34} units

For an electron the centripetal force is supplied by the Coulomb's force between protons and the electron.

We know that the angular momentum of an electron in a Bohr's orbit = integral multiple of  h'.
L = m v r = n h'
r = n h' / ( m v )
By substituting the value of v in this,  we get the expression for  radius r:

we have for our exercise,
n = 2  for the second Bohr's orbit
Z = 2  for  a He+ ion

calculating radius we get     r = 4 * 5.29 * 10^{-11} meters
calculating  angular frequency directly using the above formula:

v = linear velocity of electron in the orbit for n =2,  is    4.3 *10^6 m/sec.

So the electron has a linear speed  about 1/70th of the speed of light.

click on thanks button above please
2015-05-17T10:44:31+05:30
As it is a monoelectronic system so  just use  bohrs quantisation formula
mvr = nh/2π
here m = mass of electron
v = velocity of electron
n = principal quantum no. of the orbit
h = plancks const.
r = radius of the orbit which u can get by using the formula r = n²/z × 0.53 ×10∧-10 metre where z = the atomic no.
putting the values of m,r,n,h,2π get the value of v.
now u know v = ωr so get the ω which is asked in ur question