Answer :
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.
[tex]\frac{m v^2 }{r} = k_e\frac{Ze*e}{r^2}\\\\So\ v = r w = \sqrt{\frac{Z k_e e^2}{m r}}\\\\So\ w = \sqrt{\frac{k_e Z e^2}{m r^3}}[/tex]
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:
[tex]r = \frac{n^2 h'^2}{Z k_e e^2 m}\\\\m r * v = n h'\\\\v = \frac{Z k_e e^2}{n h'}\\\\Hence,\ r = \frac{n h'}{m v} = \frac{n^2 h'^2}{m Z k_e e^2}\\\\Finally,\ w = v / r = \frac{m Z^2 k_e^2 * e^4}{n^3 h'^3}[/tex]
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:
[tex]w = \frac{9.1 * 10^{-31} * 2^2 * (9 * 10^9)^2 * (1.6 * 10^{-19})^4}{2^3 * 1.054^3 * 10^{-34*3}}\\\\= \frac{9.1 * 4 * 81 * 1.6^4 * 10^{-89}}{8 * 1.054^3 * 10^{-102}}\\\\= 2, 062.78 * 10^{13} rad / sec\\\\= 2.063 * 10^{16} rad / sec[/tex]
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.
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.
[tex]\frac{m v^2 }{r} = k_e\frac{Ze*e}{r^2}\\\\So\ v = r w = \sqrt{\frac{Z k_e e^2}{m r}}\\\\So\ w = \sqrt{\frac{k_e Z e^2}{m r^3}}[/tex]
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:
[tex]r = \frac{n^2 h'^2}{Z k_e e^2 m}\\\\m r * v = n h'\\\\v = \frac{Z k_e e^2}{n h'}\\\\Hence,\ r = \frac{n h'}{m v} = \frac{n^2 h'^2}{m Z k_e e^2}\\\\Finally,\ w = v / r = \frac{m Z^2 k_e^2 * e^4}{n^3 h'^3}[/tex]
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:
[tex]w = \frac{9.1 * 10^{-31} * 2^2 * (9 * 10^9)^2 * (1.6 * 10^{-19})^4}{2^3 * 1.054^3 * 10^{-34*3}}\\\\= \frac{9.1 * 4 * 81 * 1.6^4 * 10^{-89}}{8 * 1.054^3 * 10^{-102}}\\\\= 2, 062.78 * 10^{13} rad / sec\\\\= 2.063 * 10^{16} rad / sec[/tex]
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.
as it is a monoelectronic system so just use bohr`s quantisation formula
mvr = nh/2π
here m = mass of electron
v = velocity of electron
n = principal quantum no. of the orbit
h = planck`s 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
mvr = nh/2π
here m = mass of electron
v = velocity of electron
n = principal quantum no. of the orbit
h = planck`s 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