# chapter 02 : Quantum Mechanics¶

## Ex2.1 : Pg:4¶

In [65]:
#Threshold wavelength of tungsten
phi = 4.5*1.6e-019     # Work function for tungesten, joule
h = 6.6e-034     # Planck's constant, Js
c = 3e+08     # Speed of light, m/s
# As phi = h*c/L0, solving for L0
L0 = h*c/phi     # Threshold wavelength of tungesten, m
print "The threshold wavelength of tungesten = %4d angstrom"%   (L0/1e-010)

The threshold wavelength of tungesten = 2750 angstrom


## Ex2.2 : Pg:44¶

In [66]:
from math import sqrt
#Maximum velocity of photoelectrons
phi = 4*1.6e-019     # Work function for photoelectric surface, joule
h = 6.6e-034     # Planck's constant, Js
e = 1.6e-019     # Electronic charge, coulomb
m = 9.1e-031     # Mass of the electron, kg
f = 1e+15     # Frequency of incident photons, Hz
c = 3e+08     # Speed of light, m/s
# KE = 1/2*m*v**2 = h*f - phi, solving for v, we have
v = sqrt(2*(h*f - phi)/m)     # Maximum velocity of photoelectrons, m/s
print "The maximum velocity of photoelectrons = %5.3e m/s" %v

The maximum velocity of photoelectrons = 2.097e+05 m/s


## Ex2.3 : Pg:45¶

In [67]:
#Energy of photoelectrons
h = 6.6e-034     # Planck's constant, Js
c = 3e+08     # Speed of light, m/s
e = 1.6e-019     # Energy equivalent of 1 joule, joule/eV
L = 1800e-010     # Wavelength of incident light, m
L0 = 2300e-010     # Threshold wavelength of tungsten, m
E = h*c*(1/L - 1/L0)     # Energy of photoelectrons emitted from tungsten, joule
print "The energy of photoelectrons emitted from tungsten = %3.1f eV"%(E/e)

The energy of photoelectrons emitted from tungsten = 1.5 eV


## Ex2.4 : Pg:45¶

In [68]:
#Longest wavelength of incident radiation
h = 6.624e-034     # Planck's constant, Js
c = 3e+08     # Speed of light, m/s
e = 1.6e-019     # Energy equivalent of 1 eV, joule/eV
phi = 6*e        # Work function of metal, joule
f0 = phi/h     # Threshold frequency for metal surface, Hz
L0 = c/f0      # Threshold (Longest) wavelength for metal, m
print "The longest wavelength of incident radiation = %4d angstrom"%(L0/1e-010)

The longest wavelength of incident radiation = 2070 angstrom


## Ex2.5 : Pg:46¶

In [69]:
#Threshold frequency and wavelength
h = 6.62e-034     # Planck's constant, Js
phi = 3.31e-019        # Work function of metal, joule
c = 3e+08     # Speed of light, m/s
f0 = phi/h     # Threshold frequency for metal surface, Hz
L0 = c/f0      # Threshold wavelength for metal, m
print "The threshold frequency for metal = %1.0e Hz"% f0
print "The threshold wavelength for metal = %4d angstrom"% round(L0/1e-10)

The threshold frequency for metal = 5e+14 Hz
The threshold wavelength for metal = 6000 angstrom


## Ex2.6 : Pg:46¶

In [70]:
from math import sqrt
#Maximum velocity of emitted electrons
h = 6.624e-034     # Planck's constant, Js
c = 3e+08     # Speed of light, m/s
m = 9.1e-031     # Mass of an electron, kg
e = 1.6e-019     # Energy equivalent of 1 eV, joule/eV
L = 4300e-010     # Wavelength of incident light, m
phi = 5*e        # Work function of nickel surface, joule
f0 = phi/h     # Threshold frequency for nickel, Hz
L0 = c/f0      # Threshold wavelength for nickel, m
print "The threshold wavelength for nickel = %4d angstrom"% (L0/1e-10)
print "Since %4d A < %4d A, the electrons will not be emitted."% (L0/1e-010, L/1e-010)
phi = 2.83*e    # Work function of potassium surface, joule
f0 = phi/h     # Threshold frequency for potassium, Hz
L0 = c/f0      # Threshold wavelength for potassium, m
print "The threshold wavelength for potassium = %4d angstrom"%(L0/1e-10)
print "Since %4d A > %4d A, the electrons will be emitted."%(L0/1e-010, L/1e-010)
# Now KE = 1/2*m*v0**2 = h*f - h*f0, where v0 is the maximum velocity
# solving for v0, we have
v0 = sqrt(2*h*c/m*(1/L - 1/L0))     # Maximum velocity of photoelectrons, m/s
print "The maximum velocity of photoelectrons = %5.3e m/s"% v0

The threshold wavelength for nickel = 2484 angstrom
Since 2484 A < 4300 A, the electrons will not be emitted.
The threshold wavelength for potassium = 4388 angstrom
Since 4388 A > 4300 A, the electrons will be emitted.
The maximum velocity of photoelectrons = 1.433e+05 m/s


## Ex2.7 : Pg:47¶

In [71]:
#Maximum energy of ejected electrons
h = 6.6e-034     # Planck's constant, Js
c = 3e+08     # Speed of light, m/s
L = 2537e-010     # Wavelength of incident light, m
L0 = 3250e-010     # Threshold wavelength of silver, m
# As U = h*(f - f0), the kinetic energy of ejected electrons
U = h*c*(1/L - 1/L0)     # Maximum energy of ejected electrons, J
print "The maximum energy of ejected electrons = %5.3e J"% U

The maximum energy of ejected electrons = 1.712e-19 J


## Ex2.8 : Pg:47¶

In [72]:
#Maximum kinetic energy and stopping potential of ejected electrons
h = 6.624e-034     # Planck's constant, Js
c = 3e+08     # Speed of light, m/s
e = 1.6e-019     # Energy equivalent of 1 eV, joule/eV
phi_0 = 1.51*e    # Work function of the metal surface, J
L = 4000e-010     # Wavelength of incident light, m
f = c/L     # Frequency of incident light, Hz
U = h*f - phi_0     # Maximum kinetic energy of ejected electrons, J
V = U/e     # Stopping potential for ejected electrons, volt
print "The maximum energy of ejected electrons = %5.3f eV"%(U/e)
print "The stopping potential of ejected electrons = %5.3f V"%(V)

The maximum energy of ejected electrons = 1.595 eV
The stopping potential of ejected electrons = 1.595 V


## Ex2.9 : Pg:48¶

In [73]:
#Work function of metal
h = 6.624e-034     # Planck's constant, Js
c = 3e+08     # Speed of light, m/s
e = 1.6e-019     # Energy equivalent of 1 eV, joule/eV
V = 1     # Stopping potential for the electrons emitted from the metal, V
L = 2500e-010     # Wavelength of incident light, m
f = c/L     # Frequency of incident light, Hz
# Now KE = h*f - phi = e*V, Einstein's Photoelectric equation, solving for phi
phi = h*f - e*V     # Work function of metal
print "The work function of metal = %5.3f eV"% (phi/e)

The work function of metal = 3.968 eV


## Ex2.10 : Pg:48¶

In [74]:
#Energy of electrons emitted from the surface of tungsten
h = 6.624e-034     # Planck's constant, Js
c = 3e+08     # Speed of light, m/s
e = 1.6e-019     # Energy equivalent of 1 eV, joule/eV
L = 1800e-010     # Wavelength of incident light, m
L0 = 2300e-010     # Threshold wavelength of tungsten, m
E = h*c*(1/L - 1/L0)     # Einstein's photoelectric equation for kinetic energy of emitted electrons, J
print "The energy of electrons emitted from the surface of tungsten = %3.1f eV"% (E/e)

The energy of electrons emitted from the surface of tungsten = 1.5 eV


## Ex2.11 : Pg:49¶

In [75]:
#Energy of photon
h = 6.624e-034     # Planck's constant, Js
c = 3e+08     # Speed of light, m/s
e = 1.6e-019     # Energy equivalent of 1 eV, joule/eV
L = 1800e-010     # Wavelength of incident light, m
L0 = 2300e-010     # Threshold wavelength of tungsten, m
E = h*c*(1/L - 1/L0)     # Einstein's photoelectric equation for kinetic energy of emitted electrons, J
print "The energy of electrons emitted from the surface of tungsten = %3.1f eV"% (E/e)

The energy of electrons emitted from the surface of tungsten = 1.5 eV


## Ex2.12 : Pg:49¶

In [76]:
from math import sqrt
#Velocity of the emitted electron
m = 9.1e-031     # Mass of electron, kg
c = 3e+08     # Speed of light, m/s
e = 1.6e-019     # Energy equivalent of 1 eV, joule/eV
phi = 2.3*e     # Work function of metal, J
L = 4300e-010     # Wavelength of incident light, m
# As 1/2*m*v**2 = h*f - phi = h*c/L - phi, Einstein's photoelectric equation
# Solving for v
v = sqrt(2*(h*c/L - phi)/m)     # Velocity of emitted electron, m/s
print "The velocity of emitted electron = %4.2e eV"% v

The velocity of emitted electron = 4.55e+05 eV


## Ex2.13 : Pg:50¶

In [77]:
#Energy of a quantum of light
c = 3e+08     # Speed of light, m/s
e = 1.6e-019     # Energy equivalent of 1 eV, joule/eV
L = 5.3e-07     # Wavelength of incident light, m
E = h*c/L     # Energy of the incident light, J
print "The energy of incident light = %4.2f eV"% (E/e)

The energy of incident light = 2.34 eV


## Ex2.14 : Pg:54¶

In [78]:
#Ratio of masses of a proton and an electron
RH = 1.09678e+07     # Rydberg constant for hydrogen, per metre
RHe = 1.09722e+07     # Rydberg constant for helium, per metre
MH_m_ratio = (RH - 1.0/4*RHe)/(RHe - RH)     # Ratio of mass of a proton to that of an electron
print "The ratio of mass of a proton to that of an electron = %4d" %MH_m_ratio

The ratio of mass of a proton to that of an electron = 1869


## Ex2.15 : Pg:56¶

In [79]:
from math import pi
#First Bohr Orbit in hydrogen atoms
n = 1     # Principle quantum number of first orbit in H-atom
h = 6.624e-034     # Planck's Constant, Js
c = 3e+08     # Speed of light, m/s
epsilon_0 = 8.85e-012     # Absolute electrical permittivity of free space, coulomb square per newton per metre square
Z = 1     # Atomic number of hydrogen
m = 9.1e-031     # Mass of an electron, kg
e = 1.6e-019     # Charge on an electron, coulomb
r = epsilon_0*n**2*h**2/(pi*m*Z*e**2)     # Radius of first Bohr's orbit, m
v = Z*e**2/(2*8.85e-012*h*n)     # Velocity of electron in the first Bohr orbit, m/s
print "The radius of first Bohr orbit = %5.3f angstrom" %(r/1e-010)
print "The velocity of electron in first Bohr orbit = (1/%3d)c"%(1/v*c)

The radius of first Bohr orbit = 0.531 angstrom
The velocity of electron in first Bohr orbit = (1/137)c


## Ex2.16 : Pg:57¶

In [80]:
#Wavelength of Balmer H_beta lines
L_Hb = 6563e-010     # Wavelength of H_beta line, m
R = 1.097e+07     # Rydberg constant, per metre
L1 = 36/(5*R)     # Wavenumber of H_alpha line, per metre
L2 = 16/(3*R)     # Wavenumber of H_beta line, per metre
L_ratio = L2/L1     # Ratio of wavelengths of H_beta and H_alpha lines
L2 = L_ratio*L1     # Wavelength of Balmer H_beta line, m
print "The wavelength of Balmer H_beta line = %4d angstrom"%(L2/1e-010)

The wavelength of Balmer H_beta line = 4861 angstrom


## Ex2.17 : Pg: 58¶

In [81]:
#First excitation energy of hydrogen atoms
n1 = 1.0     # Principle quantum number of first orbit in H-atom
n2 = 2.0     # Principle quantum number of second orbit in H-atom
m = 9.1e-031     # Mass of the electron, C
e = 1.6e-019     # Charge on an electron, coulomb
h = 6.624e-034     # Planck's Constant, Js
epsilon_0 = 8.85e-012     # Absolute electrical permittivity of free space, coulomb square per newton per metre square
U = m*e**4/(8*epsilon_0**2*h**2)*(1/n1**2 - 1/n2**2)     # First excitation energy of hydrogen atom, J
print "The first excitation energy of hydrogen atom = %5.2f eV"%(U/e)

The first excitation energy of hydrogen atom = 10.17 eV


## Ex2.18 : Pg:58¶

In [82]:
#Energy difference in the emission or absorption of sodium D1 lines
h = 6.624e-034     # Planck's Constant, Js
c = 3e+08     # Speed of light, m/s
L = 590e-09     # Wavelenght of sodium D1 line, m
E = h*c/L     # Energy difference in the emission or absorption of sodium D1 line, J
print "The energy difference in the emission or absorption of sodium D1 line = %4.2e J"% E

The energy difference in the emission or absorption of sodium D1 line = 3.37e-19 J


## Ex2.19 : Pg:58¶

In [83]:
#Wavelength of first line of Balmer series
n1 = 2.0     # Ground level of Balmer line in H-atom
n2 = 4.0     # Third level of Balmer line in H-atom
R = 1.097e+07     # Rydberg constant, per metre
L2 = 1.0/((1.0/n1**2 - 1.0/n2**2)*R)     # Wavelength of second line of Balmer series, m
n2 = 3     # Second level of Balmer line in H-atom
L1 = 1/((1.0/n1**2 - 1.0/n2**2)*R)     # Wavelength of first line of Balmer series, m
L_ratio = L1/L2     # Wavelength ratio of first and second line of Balmer series, m
L2 = 4861     # Given wavelength of second line of Balmer series, angstrom
L1 = L2*L_ratio     # Wavelength of first line of Balmer series, angstrom
print "The wavelength of first line of Balmer series = %4d angstrom" %L1

The wavelength of first line of Balmer series = 6562 angstrom


## Ex2.20 : Pg:59¶

In [84]:
#Minimum energy of the electrons in Balmer series
n1 = 2.0     # Ground level of Balmer line in H-atom
n2 = 3.0     # Second level of Balmer line in H-atom
m = 9.1e-031     # Mass of the electron, C
e = 1.6e-019     # Charge on an electron, coulomb
h = 6.624e-034     # Planck's Constant, Js
epsilon_0 = 8.85e-012     # Absolute electrical permittivity of free space, coulomb square per newton per metre square
E = m*e**4/(8*epsilon_0**2*h**2)*(1/n1**2 - 1/n2**2) # Min energy required by electron to correspond to 1st wavenumber of Balmer series, J
print "Minimum energy required by an electron to correspond to first wavenumber of Balmer series = %4.2f"%(E/e)

Minimum energy required by an electron to correspond to first wavenumber of Balmer series = 1.88


## Ex2.21 : Pg:59¶

In [85]:
#Ionization potential of hydrogen atom
m = 9.1e-031     # Mass of the electron, C
e = 1.6e-019     # Charge on an electron, coulomb
h = 6.626e-034     # Planck's Constant, Js
epsilon_0 = 8.85e-012     # Absolute electrical permittivity of free space, coulomb square per newton per metre square
phi = m*e**4/(8*epsilon_0**2*h**2)     # Work function or ionization energy of hydrogen atom, J
print "The ionization energy of hydrogen atom = %5.2f eV"%(phi/e)

The ionization energy of hydrogen atom = 13.55 eV


## Ex2.22 : Pg:60¶

In [86]:
#Wavelength of second number of Balmer series of hydrogen
n1 = 2.0     # Principle quantum number of second orbit in H-atom
n2 = 3.0     # Principle quantum number of third orbit in H-atom
R = 1.097e+07     # Rydberg constant, per metre
L1 = 1/((1/n1**2 - 1/n2**2)*R)     # Wavelength of first Balmer line, m
n2 = 4.0     # Principle quantum number of third orbit in H-atom
L2 = 1/((1.0/n1**2 - 1/n2**2)*R)     # Wavelength of second Balmer line, m
L_ratio = L2/L1     # Wavelength ratio of second and first line of Balmer series
L1 = 6563e-010     # Given wavelength of first line of Balmer series, m
L2 = L_ratio*L1     # Wavelength of second Balmer line, m
print "The wavelength of second Balmer line = %4e m"%(L2)

The wavelength of second Balmer line = 4.861481e-07 m


## Ex2.23 : Pg:60¶

In [87]:
#Wavelength of emitted light
e = 1.6e-019     # Charge on an electron, coulomb
h = 6.624e-034     # Planck's Constant, Js
n = 2     # Principal quantum number for second orbit in H-atom
V = 13.6     # Ionization potential of H-atom, V
U1 = -1*V*e     # Energy of electron in first orbit, J
U2 = U1/n**2     # Energy of electron in second orbit, J
# As U2 - U1 = h*c/L, solving for L
L = h*c/(U2 - U1)     # Wavelength of light emitted in the transition from second orbit to the first orbit, m
print "The wavelength of light emitted in the transition from second orbit to the first orbit = %4d angstrom"%(L/1e-010)

The wavelength of light emitted in the transition from second orbit to the first orbit = 1217 angstrom


## Ex2.24 : Pg:61¶

In [88]:
#Radius and speed of electron in the first Bohr orbit
m = 9.1e-031     # Mass of the electron, C
e = 1.6e-019     # Charge on an electron, coulomb
h = 6.626e-034     # Planck's Constant, Js
epsilon_0 = 8.85e-012     # Absolute electrical permittivity of free space, coulomb square per newton per metre square
Z = 1;n = 1
r_H = epsilon_0*n**2*h**2/(pi*m*Z*e**2)     # Radius of first Bohr orbit, m
v_H = Z*e**2/(2*epsilon_0*n*h)     # Velocity of the electron in the first Bohr orbit, m/s
print "The radius of first Bohr orbit = %4.2e m"%(r_H)
print "The velocity of the electron in the first Bohr orbit = %3.1e m/s"%(v_H)

The radius of first Bohr orbit = 5.31e-11 m
The velocity of the electron in the first Bohr orbit = 2.2e+06 m/s


## Ex2.25 : Pg:61¶

In [89]:
#Radius and velocity of electron for H and He
m = 9.1e-031     # Mass of the electron, kg
e = 1.6e-019     # Charge on an electron, coulomb
h = 6.624e-034     # Planck's Constant, Js
epsilon_0 = 8.85e-012     # Absolute electrical permittivity of free space, coulomb square per newton per metre square
c = 3e+08     # Speed of light, m/s
Z = 1; n = 1     # Atomic number and principal quantum number of H-atom
r_H = epsilon_0*n**2*h**2/(pi*m*Z*e**2)     # Radius of first Bohr orbit for H-atom, m
v_H = Z*e**2/(2*epsilon_0*n*h)     # Velocity of the electron in the first Bohr orbit of H-atom, m/s
print "The radius of first Bohr orbit = %4.2e m"%(r_H)
print "The velocity of the electron in the first Bohr orbit = %3.1e m/s"%(v_H)
print "The velocity of the electron in H-atom compared to the velocity of light = %4.2e"%(v_H/c)
Z = 2     # Atomic number of He-atom
r_He = r_H/Z     # Radius of first Bohr orbit for He-atom, m
v_He = 2*v_H     # Velocity of the electron in the first Bohr orbit of He-atom, m/s
print "The radius of first Bohr orbit = %4.2e m"%(r_He)
print "The velocity of the electron in the first Bohr orbit = %3.1e m/s"%(v_He)
print "The velocity of the electron in He-atom compared to the velocity of light = %5.3e"%(v_He/c)

The radius of first Bohr orbit = 5.31e-11 m
The velocity of the electron in the first Bohr orbit = 2.2e+06 m/s
The velocity of the electron in H-atom compared to the velocity of light = 7.28e-03
The radius of first Bohr orbit = 2.65e-11 m
The velocity of the electron in the first Bohr orbit = 4.4e+06 m/s
The velocity of the electron in He-atom compared to the velocity of light = 1.456e-02


## Ex2.26 : Pg:62¶

In [90]:
#Difference in wavelength in the spectra of hydrogen and deuterium
R_H = 1.097e+07     # Rydberg constant for H-atom, per metre
M_H = 1.0     # Mass of H-atom, amu
M_D = 2*M_H     # Mass of D-atom, amu
m = 0.000549*M_H     # Mass of an electron, amu
R_D = R_H*(1+m/M_H)/(1+m/M_D)     # Rydberg constant for D-atom, per metre
n1 = 2.0; n2 = 3.0     # Principal qunatum numbers for first line of Balmer series
L_H = 1/(R_H*(1/n1**2 - 1/n2**2))     # Wavelength of H-atom, m
L_D = 1/(R_D*(1/n1**2 - 1/n2**2))     # Wavelength of D-atom, m
delta_H = (L_H - L_D)/1e-010     # Difference in wavelength in the spectra of hydrogen and deuterium, angstrom
print "The difference in wavelength in the spectra of hydrogen and deuterium = %3.1f angstrom"% delta_H

The difference in wavelength in the spectra of hydrogen and deuterium = 1.8 angstrom


## Ex2.27 : Pg:63¶

In [91]:
#Ionization energy of hydrogen atom with orbiting muon
m = 9.1e-031     # Mass of the electron, kg
e = 1.6e-019     # Charge on an electron, coulomb
h = 6.624e-034     # Planck's Constant, Js
epsilon_0 = 8.85e-012     # Absolute electrical permittivity of free space, coulomb square per newton per metre square
m1 = 200*m     # Mass of muon, kg
phi1 = m1*e**4/(8*epsilon_0**2*h**2)     # Ionization energy of H-atom with muon, J
print "The ionization energy of hydrogen atom with orbiting muon = %4.2e eV"%(phi1/1.6e-019)
# Result
# The ionization energy of hydrogen atom with orbiting muon = 2.71e+003 eV

The ionization energy of hydrogen atom with orbiting muon = 2.71e+03 eV


## Ex2.28 : Pg:64¶

In [92]:
#Photon emitted by hydrogen atom
e = 1.6e-019     # Energy equivalent of 1 eV, joule/eV
h = 6.624e-034     # Planck's constant, Js
c = 3e+08     # Speed of light, m/s
E1 = -13.6     # Energy of electron in the first orbit of hydrogen atom, eV
n = 2     # Principal quantum number for second orbit
E2 = E1/n**2     # Energy of electron in the second orbit of hydrogen atom, eV
E = (E2 - E1)*e     # Energy of photon emitted, joule
P = E/c     # Momentum of photon, kg-m/s
L = (h/P)/1e-010     # de_Broglie wavelength of photon, angstrom
print "The energy of photon emitted by hydrogen atom %5.2e J"% E
print "The momentum of photon = %4.2e kg-m/s"% P
print "The de_Broglie wavelength of photon = %4d angstrom"% L

The energy of photon emitted by hydrogen atom 1.63e-18 J
The momentum of photon = 5.44e-27 kg-m/s
The de_Broglie wavelength of photon = 1217 angstrom


## Ex2.29 : Pg:64¶

In [93]:
#Energy required to create a vacancy in Cu
n = 1     # Principal quantum number of K shell
Z = 29     # Atomic number of copper
U = 13.6     # Ionization potential of hydrogen atom, eV
E1 = Z**2*U/n**2     # Energy required to create a vacancy in K-shell of copper atom, eV
n = 2     # Principal quantum number of L shell
E2 = Z**2*U/n**2     # Energy required to create a vacancy in K-shell of copper atom, eV
print "The energy required to create a vacancy in K-shell of copper atom = %5.2e eV"% E1
print "The energy required to create a vacancy in L-shell of copper atom = %5.2e eV"% E2

The energy required to create a vacancy in K-shell of copper atom = 1.14e+04 eV
The energy required to create a vacancy in L-shell of copper atom = 2.86e+03 eV


## Ex2.30 : Pg:65¶

In [94]:
#Excitation potential for mercury
e = 1.6e-019     # Energy equivalent of 1 eV, joule/eV
h = 6.624e-034     # Planck's constant, Js
c = 3e+08     # Speed of light, m/s
L = 2537e-010     # Wavelength of absorbed line of Hg, m
V = h*c/(e*L)     # Excitation potential for Hg, v
print "The excitation potential for Hg = %3.1f V"% V

The excitation potential for Hg = 4.9 V


## Ex2.31 : Pg:65¶

In [95]:
#Atomic number of impurity in Zinc target
L1 = 1.43603e-010     # Wavelength of characteristic K_alpha line from Zn, m
Z1 = 30.0     # Atomic number of zinc
L2 = 0.53832e-010     # Wavelength of unknown line from Zn, m
# As (1/L1)/(1/L2) = (Z1/Z2)**2, solving for Z2
Z2 = Z1*(L1/L2)**(1.0/2)     # Atomic number of impurity in Zn target
print "The atomic number of impurity in Zn target = %2d"% round(Z2)

The atomic number of impurity in Zn target = 49


## Ex2.32 : Pg:65¶

In [96]:
#Mu-mesonic atom subjected to Bohr orbit
Z = 3     # Atomic number of Mu-mesonic atom
m_e = 9.1e-031     # Mass of the electron, kg
e = 1.6e-019     # Charge on an electron, coulomb
h = 6.624e-034     # Planck's Constant, Js
epsilon_0 = 8.85e-012     # Absolute electrical permittivity of free space, coulomb square per newton per metre square
m = 200*m_e     # Mass of a muon, kg
# As r_H = epsilon_0***h**2/(pi*m*(e**2) and r = epsilon_0*n**2*h**2/(pi*m*Z*(e**2)
# r = r_H gives
n = sqrt(m/m_e*Z)     # Value of n for which r = r_H
n1 = 1.0; n2 = 2.0     # Principal quantum numbers corresponding to first excitation
U = m*e**4*Z**2.0/(8*epsilon_0**2*h**2*1.6e-019)*(1/n1**2-1/n2**2)     # First excitation potential of the atom, eV
print "The value of n for which radius of orbit is equal to Bohr radius = %2d"% round(n)
print "The first excitation potential of the atom = %4.2e eV"% U

The value of n for which radius of orbit is equal to Bohr radius = 24
The first excitation potential of the atom = 1.83e+04 eV