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(solution) BOHR'S ATOMIC MODEL In Bohr's model of the atom, two key


derive the formulas by using the steps on page 3 and 4 form the attached file.


BOHR?S ATOMIC MODEL

 

In Bohr?s model of the atom, two key assumptions must be made that go against the predictions of classical

 

electromagnetism, as encapsulated by Maxwell?s theory. I?ll state them here, and we will explore them more in

 

the following pages.

 

? Electrons orbit the nucleus of the atom in stable circular or elliptical orbits without losing energy (classical electromagnetism predicts that orbiting electrons, because they are constantly accelerating, emit

 

electromagnetic radiation and so lose energy continuously, leading them to spiral into the nucleus.)

 

? Electrons can only lose or gain energy by emitting radiation in certain chunks given by Planck?s formula,

 

and in doing so the electrons instantaneously jump to a different orbit with an energy that is higher

 

or lower by the amount radiated. The radiation is emitted at a single frequency according to Planck?s

 

formula, and that frequency is determined by the difference in energy between the orbit the electron starts

 

out in and the orbit the electron ends up in. (Classical electromagnetism says that the emitted radiation

 

should be determined by the frequency of a single orbit, not by the difference between two orbits, and

 

it also predicts that electrons should spiral between orbits, losing energy, and emitting radiation over a

 

continuous range of frequencies.)

 

These two assumptions may be taken to be principles of a new physical theory (which we now know to be

 

quantum theory!). If we accept them, then we can get an idea of what they predict for potentially observable

 

phenomena using basic mechanics, which we will do below.

 

The two principles imply that the motion of the electrons in their orbits can be treated classically as particles

 

in stable orbits similar to planets orbiting the Sun. Let?s assume for simplicity that the orbit is circular. In that

 

case, the energy of an electron in a circular orbit of radius r and orbital speed v is just the sum of its kinetic

 

energy an potential energy:

 

E= ke2

 

1

 

mv 2 ?

 

,

 

2

 

r k= 1

 

4?0 (1) where the electric potential between two point charges q1 = +e (the charge of the nucleus of Hydrogen) and

 

1 q1 q2

 

q2 = ?e (the charge of the electron) a distance r apart comes just from Coulomb?s law U = 4?

 

r where 0

 

0

 

is the permittivity of free space. m is the mass of the electron.

 

The magnitude of the electron?s angular momentum L is defined by L = mvr, which can be rearranged to give:

 

v= L

 

mr (2) so substituting this into equation (1) we get:

 

L2

 

ke2

 

?

 

.

 

(3)

 

2

 

2mr

 

r

 

Now, the electron is moving in its orbit according to Newton?s second law of motion F = ma where F is the

 

magnitude of the force between the electron and the central nucleus, given by Coulomb?s law:

 

E= F = ke2

 

1 q1 q2

 

= 2

 

2

 

4?0 r

 

r (4) and the acceleration of the electron in circular motion is given by a = v 2 /r. Then,

 

F = ma ? mv 2

 

ke2

 

= 2

 

r

 

r ? 1

 

ke2 m

 

=

 

r

 

L2 (5) Substituting L for v in the above equation, we get:

 

L2

 

ke2

 

=

 

mr3

 

r2 ? r= L2

 

kme2 1 ? 1

 

k 2 e4 m 2

 

=

 

r2

 

L4 (6) Substituting these expressions into equation (3).

 

L k 2 e4 m 2

 

ke2 m

 

k 2 e4 m k 2 e4 m

 

1 k 2 e4 m

 

? ke2 2 =

 

?

 

=?

 

.

 

(7)

 

4

 

2

 

2

 

2m L

 

L

 

2L

 

L

 

2 L2

 

So we end up with the result that, for an electron in a circular orbit with an angular momentum L, its energy

 

E is

 

E= 1 k 2 e4 m

 

.

 

(8)

 

2 L2

 

Note that, apart from L, all other quantities on the right hand side of the equation are constants. The fact that

 

its energy is negative means the electron is bound to the Hydrogen atom.

 

Therefore, when an electron makes a transition between an upper and a lower orbit (higher and lower energy),

 

the change in energy ?E of the electron is given by

 



 



 

1

 

k 2 e4 m

 

1

 

?

 

.

 

(9)

 

Eupper ? Elower = ?E =

 

2

 

L2lower

 

L2upper

 

E=? According to Bohr?s model of the atom, this energy is emitted as a light, of a single frequency given by Planck?s

 

formula Eemitted = h?. So setting the change in energy of the electron jumping from the upper to lower orbit

 

to the energy of the emitted photon:

 

? ?E = Eemitted = h? ?= ?E

 

h (10) The wavelength of light emitted is given by ? = c/?, so

 

1

 

?E

 

=

 

.

 

?

 

ch

 

Using Balmer?s empirical formula for the wavelengths of emission lines from Hydrogen:

 



 



 

1

 

1

 

1

 

=R

 

?

 

nupper > nlower

 

?

 

n2lower

 

n2upper (11) (12) we get

 



 

?E = chR 1

 

n2lower ? 1

 

n2upper 

 

(13) Where R is Rydberg?s constant, which is determined by spectroscopy. The theory (equation 9) must match

 

experiment (equation 12 and thus 13), and it does so if L must be proportional to an integer: L ? n. In order

 

that the equations match in detail (i.e. equation (9) predicts the correct value for R), L must be

 

h

 

= n~

 

(14)

 

2?

 

where we?ve defined ~ = h/2?. Therefore the electron angular momentum is quantized - that is, it can only take

 

discrete, integer values of Planck?s constant divided by 2?, unlike a classical particle which can take any value

 

of angular momentum it can!.

 

L=n 2 HOMEWORK: DEMONSTRATION OF THE CORRESPONDENCE

 

PRINCIPLE

 

One important requirement of a new physical theory is it must reproduce all of the experimental results that

 

the old theory does. The simplest way to do this is to say that the old theory is NOT incorrect - it still

 

applies within a certain domain - and that the new theory is simply an extension/modification of the old theory

 

into a new physical domain. For example, quantum theory does not render classical (Newtonian/Maxwellian)

 

mechanics and electromagnetic theories incorrect - it simply modifies them to work in the domain of the very

 

very small. That means that, as one moves from the very very small towards the not-so-small, quantum theory

 

must start looking more and more like classical theory. This is the essence of the correspondence principle

 

in Bohr?s model. In Bohr?s model, whether you are dealing with the?very, very small? or the ?not-so-small?

 

depends on the quantum number n. For small n, we are dealing with small electron orbits close to the atom?s

 

nucleus. For very large n we are dealing with much larger orbits. Therefore, as n becomes very large n ? ?,

 

Bohr?s model must end up giving us just what Maxwell?s electrodynamics gives us.

 

What does Maxwell?s electrodynamics give us? It says that the frequency of the emitted radiation from the

 

atom ?emitted must be equal to the electron?s orbital frequency ?orbit :

 

MAXWELL : ?emitted = ?orbit (15) Bohr?s model says, on the other hand:

 

?E

 

(16)

 

h

 

where ?E is given by equation (13). The correspondence principle says that as nlower , nupper become very large

 

in equation (13), equations (15) and (16) must become equivalent to each other:

 



 



 

?E

 

1

 

1

 

= cR

 

?

 

? ?orbit

 

nlower , nupper ? ?

 

(17)

 

h

 

n2lower

 

n2upper

 

BOHR : ?emitted = Let?s show this! Complete the following steps:

 

1. Let?s take the case where the electron is jumping between two adjacent orbits - i.e. that nlower = n

 

nupper = n + 1, where n is an integer, so that

 



 



 

1

 

1

 

?E

 

= cR

 

?

 

(18)

 

h

 

n2

 

(n + 1)2

 

Show that when n becomes very large, equation (18) becomes:

 

2cR

 

?E

 

? 3

 

h

 

n (19) and by comparing equations (9) and (13) (and using equation (14)) show that

 

R= k 2 e4 m

 

4?c~3 (20) so that (19) becomes

 

?E

 

k 2 e4 m 1

 

?

 

h

 

2?~3 n3 3 (21) 2. The orbital frequency of a circular orbit is

 

?orbit = vn

 

2?rn (22) where vn is the orbital speed of the nth orbit, and rn is that orbit?s radius.

 

From equations (2) and (14) show that

 

n~

 

mrn (23) rn = n2 A (24) vn =

 

and from equation (6) show that where we define the constant A to be

 

A= ~2

 

.

 

kme2 (25) vn = ~

 

mnA (26) Therefore show that 3. The final step: Using the results from the Step 2, show that

 

?orbit = ~

 

1

 

k 2 e4 m 1

 

=

 

2?mA2 n3

 

2?~3 n3 (27) and hence show that in the limit of large n, the Bohr model prediction (16) gives the same answer as the

 

prediction from Maxwell?s theory (15) 4

 


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