Bohr Model of the Atom - MSU Chemistry - Michigan State University

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i ˆjare unit vectors. The velocity becomes. (. ) V= ˆ. ˆ sin cos d d dt dt rr i j θ θ θ. = −. +. G. G and the ma
Bohr Model of the Atom The Hydrogen atom consists of a relatively massive positively charged (e) nucleus (proton) around which an electron of charge –e moves. The force attracting the electron to the proton is given by Coulombs law

G G e2rˆ f = 4πε 0 r 2 r is the magnitude of the position vector from the proton to the electron and ε 0 = 8.85419 x10−12 C 2 N −1m−2 is the permittivity of free space. The velocity of the electron is equal to the time derivative of the position

G drG vector V= and if the orbit is circular dt

G r = r cosθ iˆ + r sin θ ˆj where iˆ & ˆj are unit vectors. The velocity becomes

G

G dr V= dt

dθ dt

=r

( − sinθ iˆ + cosθ ˆj )

and the magnitude is V=r

dθ = rω dt

We need the acceleration which is G G dV d 2 r = dt dt 2

d 2θ =r 2 dt

(

− sin θ iˆ + cosθ ˆj

)

⎛ dθ ⎞ −r⎜ ⎟ ⎝ dt ⎠

2

(cosθ iˆ + sinθ ˆj )

Identifying

G rˆ = cosθ iˆ + sin θ ˆj

(

J. F. Harrison

)

Michigan State University

1

means that the counterbalancing force is ⎛ dθ ⎞ rm ⎜ ⎟ = rmω 2 = ⎝ dt ⎠ 2

mV 2 r

Equating the two forces gives

e2

4πε 0

= rmω 2 = 2 r

mV 2 r

Or 2 =2 n2 e2 m = ( mV ) = p 2 = 2 4πε 0r r

Where Bohr used the de Broglie relationship p =

h

λ

and assumed that

2π r = nλ n = 1, 2,3, . This results in the radius of the electrons orbit being

=2n2 4πε 0 = a0 n 2 n = 1,2,3," r= 2 em

=2 4πε 0 is equal to 0.529177 x10−10 m . Note that it is the 2 em smallest radius possible for the electron ( n = 1) and is called the Bohr radius. One often refers to n = 1 as the first Bohr orbit, n = 2 as the second Bohr Where a0 =

orbit and so on. The energy of the electron in the H atom is the sum of its kinetic ( T ) and potential energies ( V )

E = T +V

e = 12 mV 2 − 4πε

2 0r

=−

e2

8πε 0 r

And using the value of the radius found above the energy becomes

J. F. Harrison

Michigan State University

2

En = −

me4

E

2 ( 4πε 0 ) =2 n 2 2

= 21 n

n = 1,2,3,"

Where E1 = −

me4

2 ( 4πε 0 ) = 2

2

= −2.17987 x10−18 J = −13.606eV

E1 is the lowest possible energy for the H atom. As with the radius of the allowed orbits one refers to E1 as the energy of the electron in the first Bohr orbit, E2 as the energy of the electron in the second Bohr orbit and so on. It’s useful to remember the possible energies of the H atom as

En = −

13.606 eV n = 1,2,3," n2

How fast is the electron moving in the allowed orbits? We can estimate this as follows Since

2π r = nλ =

nh p

We have

p=

nh nh = = mv 2π r 2π a0 n2

And so

h = = =v 2π a0 mn a0 mn So the speed (magnitude of the velocity) is v =

= 2.1877 x106 = m/s a0 mn n

It’s insightful to express this speed relative to the speed of light c.

v=

J. F. Harrison

c 137.0

Michigan State University

3