Lectuer 15

advertisement
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 15
The shape of the hydrogen atom
s Orbitals Are Spherically Symmetric
- The hydrogen atomic wave functions depend on three
quantum numbers, n, Ɩ, and m.
- The quantum number n
is called the principal quantum number and has the
values 1,2,……
- The energy depends on only the principal quantum
number through the equation En = -e2/ 8πε0a0n2.
- The quantum number Ɩ
-
is called the angular momentum quantum number and
has the values 0, 1, …….., n-1.
- The angular momentum of the electron about the proton
is determined completely by Ɩ through L= ħ√ Ɩ (Ɩ+1).
- Note that the form of the radial wave functions depends
on both n and Ɩ.
- If Ɩ = 0 being denoted by s, Ɩ = 1 by p, Ɩ = 2 by d, Ɩ = 3 by f
- The letter s, p, d, and f stand for sharp, principal, diffuse,
and fundamental.
- A wave function with n = 1 and Ɩ = 0 is called 1s wave
function; one with n =2 and Ɩ = 0 a 2s wave function, and
so on.
[1]
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 15
- The third quantum number m is called the magnetic
quantum number
- It takes on the 2 Ɩ + 1 values m = 0, ±1, ±2, ……, ± Ɩ.
- The z component of the angular momentum is
determined completely by m through L z = m ħ.
- The quantum number m is called the magnetic quantum
number because the energy of a hydrogen atom in a
magnetic field depends on m.
- The (2 Ɩ + 1) – fold degeneracy in the absence of a
magnetic field is split into 2 Ɩ + 1 different energies in the
presence of a magnetic field, and this splitting depends on
the (2 Ɩ + 1) different values that m can have.
- This splitting is illustrated in figure (1).
The complete hydrogen atomic wave functions depend on three
variables and so plot them or to display them.
- It is common to consider the radial and the angular parts
separately.
[2]
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 15
- The state of lowest energy of a hydrogen atom is the 1s
state. The radial function associated with the 1s state is
(Z = 1)
R1s (r ) 
2
 r / a0
e
3/2
a0
(1)
The radial wave functions are normalized with respect to
integration over r, and so we have that
2


4
2 2 r / a0
dr
r
R
(
r
)

dr
r
e
 1 (2)


1s
3 
0
a0 0
2
For 1s state, the probability that the electron lies between
r and r + d r
Pr ob 
4 2 2 r /a0
r e
dr
3
a0
[3]
(3)
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 15
Example1:
Calculate the probability that an electron described by a
hydrogen atomic 1s wave function will be found within
one Bohr radius of the nucleus.
Solution

Prob (0≤ r≤ a0) =
4
2 2 r / a0
dr
r
e
3 
a0 0
1
2 2 x
dx
x
e
=4 
0
=4
1
0
 2 x x 2 2x 2 
e ( 2  4  8 ) 


 1  2  1 2 2 
4(
=  4   e  2  4  8 )
 


= 1-5e-2
= 0.323
- We must keep in mind that we are dealing with only the
radial parts of the total wave function here.
- The radial parts are easy to display because they depend
on only the one coordinate r.
- The angular parts depend on both 𝛉 and 𝛗 but in 1s Ɩ =0
and m=0
[4]
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 15
2

 d   d  sin 
0
- Prob (1s) =


1s
(r , ,  ) 1s (r ,  ,  )r 2dr
0
(4)
2
4r 2 r /a0
e
dr
a03
- Eq. (4) as the same Eq.(3) because the 1s orbital is
spherically symmetric, and so the integration over 𝛉 and
𝛗 introduces no additional factors.
- We can use equation 4 to calculate average values of r. for
example

r
1s
4
3
3 2 r / a0
 3 r e
dr  a0
a0 0
2 (4)
- This equation can be used to determine the mostprobable
distance of a 1s electron from the nucleus.
[5]
Download