# Chapter 9: Multielectron Atoms - Ground states and X-Ray Excitations¶

## Example 9.4, Page 316¶

In [2]:
#Variable declaration
Z = 2.

#Calculations
#At ground state
n1=n2 =1
Eg = -(Z**2*13.6/n1**2)-(Z**2*13.6/n2**2)

#At first excited state
n1=1
n2=2
Ef = -(Z**2*13.6/n1**2)-(Z**2*13.6/n2**2)

#Result
print "Energy at ground state is %.f eV"%Eg
print "Energy at first excited state is %d eV"%Ef

Energy at ground state is -109 eV
Energy at first excited state is -68 eV


## Example 9.5, Page 325¶

In [3]:
#Variable declaration
Z=[16, 8, 3]

#Calculations&Results
#Argon numbers
n = 1
for x in range(len(Z)):
E=(-((Z[x])/n)**2)*13.6
print "The electric field for n=%d is  %.1f  ev"%(n, E)
n+=1

#Answer differnce is because of round off

The electric field for n=1 is  -3481.6  ev
The electric field for n=2 is  -217.6  ev
The electric field for n=3 is  -13.6  ev


## Example 9.7, Page 337¶

In [4]:
#Variable declaration
Z=92
n=2

#Calculations
E=((Z/n)**2)*13.6#in ev

#Result
print "The ionization energy is  %.e  ev"%E

The ionization energy is  3e+04  ev


## Example 9.8, Page 340¶

In [5]:
#Calculations&Results
#Energy of K shell
z=26
k=2
E_k=13.6*(z-k)**2#in ev
v=7.8*10**3#in V

#for L shell
l=10
E_l=13.6*(z-l)**2#in ev
h=E_k-E_l
R_m=1.1*10**7
x=R_m*(z-2)**2#x=1/lamda
lamba=1/x
print "The wavelength is  %.1e  m"%lamba

The wavelength is  1.6e-10  m


## Example 9.9, Page 342¶

In [6]:
#Variable declaration
#Energy of K shell
z=82
k=2

#Calculations
E_k=13.6*(z-k)**2#in ev

#Result
print "The energy of K shell is  %.1e  eV"%E_k

The energy of K shell is  8.7e+04  eV