# 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)

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by liz98 15.05.2015

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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 15.05.2015

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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.

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.

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