# Chapter 14: Development of Quantum Mechanics¶

## Example 14.1, Page 14.20¶

In [1]:
# Given
E = 75 # energy of photon in eV
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in J

#Calculations
f = E * e / h
lamda = c / f

#Result
print "Frequency is %.2e Hz\nWavelength is %.1f A"%(f,lamda * 10**10)


Frequency is 1.81e+16 Hz
Wavelength is 165.5 A


## Example 14.2, Page 14.20¶

In [2]:
# Given
P = 2e5 # radiated power in W
f = 98e6 # frequency in Hz
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C

#Calculations
E = h * f
n = P / E

#Result
print "Number of quanta emitted per sec is %.2e"%n


Number of quanta emitted per sec is 3.08e+30


## Example 14.3, Page 14.20¶

In [3]:
# Given
lamda = 4e-7 # wavelength of spectral line in meter
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C

#Calculations
E = (h * c) / lamda

#Result
print "Energy of photon is %.3e J"%E


Energy of photon is 4.965e-19 J


## Example 14.4, Page 14.21¶

In [7]:
# Given
lamda = 5e-7 # wavelength of green light in meter
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C
P = 1. # energy in erg

#Calculations
E = ((h * c) / lamda) * (10**7)
n = P / E

#Result
print "Number of photons of green light emitted is %.2e"%n


Number of photons of green light emitted is 2.52e+11


## Example 14.5, Page 14.21¶

In [8]:
# Given
E = 5e-19 # energy of photon in J
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C

#Calculations
lamda = (c * h) / E

#Result
print "Wavelength is %.f A"%(lamda * 10**10)

Wavelength is 3972 A


## Example 14.6, Page 14.21¶

In [9]:
# Given
lamda = 4.35e-7 # wavelength of green light in meter
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C
P = 1 # energy in erg

#Calculations
E = ((h * c) / lamda)

#Result
print "Energy of an electron is %.3e J"%E


Energy of an electron is 4.566e-19 J


## Example 14.7, Page 14.22¶

In [14]:
# Given
lamda = 5.6e-7 # wavelength of light in meter
n = 120 # no. of photons per second
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C

#Calculations
E = ((h * c) / lamda)
p = E * n

#Result
print "Energy received by the eye per second is %.3e W"%p

Energy received by the eye per second is 4.256e-17 W


## Example 14.8, Page 14.22¶

In [15]:

# Given
lamda = 5.5e-7 # wavelength of light in meter
E = 1.5 # energy in J
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C

#Calculations
E_ = ((h * c) / lamda)
n = E / E_

#Result
print "Number of photons of yellow light = %.3e"%n


Number of photons of yellow light = 4.154e+18


## Example 14.9, Page 14.22¶

In [16]:
from math import *

# Given
lamda = 4.35e-7 # wavelength of light in meter
lambda_ = 5.42e-7 # threshold wavelength of photoelectron in meter
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C
m = 9.1e-31 # mass of an electron in kg

#Calculations
w = ((h * c) / lambda_)
v = sqrt(((2 * h * c) / m) * (1 / lamda - 1 / lambda_))
V = m * v**2 / (2 * e)

#Result
print "Work function is %.3e J\nStopping potential is %.2f V\nMaximum velocity is %.3e m/sec"%(w,V,v)


Work function is 3.664e-19 J
Stopping potential is 0.56 V
Maximum velocity is 4.451e+05 m/sec


## Example 14.10, Page 14.23¶

In [17]:

# Given
f = 1.2e15 # frequency of light in Hz
f_ = 1.1e+15 # threshold frequency of photoelectron emission in copper in Hz
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C

#Calculations
E = h * (f - f_) / e

#Result
print "Maximum energy of photoelectron is %.3f eV"%E


Maximum energy of photoelectron is 0.414 eV


## Example 14.11, Page 14.23¶

In [20]:

# Given
lamda = 6.2e-7 # threshold wavelength of photoelectron in first case in meter
lambda_ = 5e-7 # threshold wavelength of photoelectron in second case in meter
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C

#Calculations
w = ((h * c) / lamda) * (1 / e)
w_ = ((h * c) / lambda_) * (1 / e)

#Result
print "Work function for wavelength %.e A is %.f eV\nWork function for wavelength %.e A is %.2f eV"%(lamda,w,lambda_,w_)


Work function for wavelength 6e-07 A is 2 eV
Work function for wavelength 5e-07 A is 2.48 eV


## Example 14.12, Page 14.24¶

In [23]:
# Given
lamda = 3.132e-7 # wavelength of light in meter
V = 1.98 # stopping potential in V
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C
m = 9.1e-31 # mass of an electron in kg

#Calculations
E = e * V
lambda_ = 1. / ((1. / lamda) - (E / (h * c)))
f = c / lambda_
w = ((h * c) / lambda_)

#Result
print "Work function is %.3e J\nMaximum energy is %.3e J\nThreshold frequency is %.3e Hz"%(w,E,f)

Work function is 3.173e-19 J
Maximum energy is 3.168e-19 J
Threshold frequency is 4.793e+14 Hz


## Example 14.13, Page 14.24¶

In [4]:
# Given
w = 4.8 # work function in eV
lambda1 = 5e-7 # wavelength of incident radiation in first case in meter
lambda2 = 2e-7 # wavelength of incident radiation in second case in meter
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C

#Calculations
E_k1 = h*c/lambda1
E_k2 = h*c / lambda2

#Result
print "The energy corresponding to wavelength 5000 A is %.2f which is found to be less than the work function 4.8 eV"%(E_k1/e)
print "The energy corresponding to wavelength 2000 A %.2f is found to be greater than the work function"%(E_k2/e)


The energy corresponding to wavelength 5000 A is 2.48 which is found to be less than the work function 4.8 eV
The energy corresponding to wavelength 2000 A 6.21 is found to be greater than the work function


## Example 14.14, Page 14.25¶

In [29]:

# Given
lamda = 5.893e-7 # wavelength of light in meter
V = 0.36 # stopping potential for emitted electron in eV
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C
m = 9.1e-31 # mass of an electron in kg

#Calculations
E = h * c / lamda
w = ((h * c) / lamda) * (1 / e) - V
f = w * e / h

#Result
print "Maximum energy is %.2f eV\nWork function is %.2f eV\nThreshold frequency is %.2e cycles/sec"%(E/e,w,f)


Maximum energy is 2.11 eV
Work function is 1.75 eV
Threshold frequency is 4.22e+14 cycles/sec


## Example 14.15, Page 14.25¶

In [30]:
# Given
lamda = 5.89e-7 # wavelength of light in meter
lambda_ = 7.32e-7 # threshold wavelength of photoelectron in meter
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C
m = 9.1e-31 # mass of an electron in kg

#Calculations
E = (h * c) * (1 / lamda - 1 / lambda_)
V = E / e

#Result
print "Stopping potential is %.3f V\nMaximum kinetic energy is %.3e J"%(V,E)


Stopping potential is 0.412 V
Maximum kinetic energy is 6.587e-20 J


## Example 14.16, Page 14.26¶

In [31]:

# Given
E = 1.5 # maximum energy in eV
lambda_ = 2.3e-7 # threshold wavelength of photoelectron in meter
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C
m = 9.1e-31 # mass of an electron in kg

#Calculations
lamda = 1 / ((E * e / (h * c)) + (1 / lambda_))

#Resuult
print "Wavelength of light is %.1f A"%(lamda * 1e10)


Wavelength of light is 1799.8 A


## Example 14.17, Page 14.26¶

In [33]:

# Given
lamda = 1.5e-7 # wavelength of light in in meter
w = 4.53 # work function of tungsten in eV
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C

#Calculations
E = ((h * c) / lamda) * (1 / e)
k = E - w

#Result
print "Energy of incident photon is %.2f eV,which is greater than the work function \nSo it causes photoelectric emission.\nKinetic energy of the emitted electron is %.2f eV"%(E,k)


Energy of incident photon is 8.28 eV,which is greater than the work function
So it causes photoelectric emission.
Kinetic energy of the emitted electron is 3.75 eV


## Example 14.18, Page 14.26¶

In [34]:

# Given
w = 2.3 # work function of sodium in eV
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C

#Calculations
lamda = ((h * c) / w) * (1 / e)

#Result
print "Longest wavelength required for photoemission is %.2f A"%(lamda * 1e10)


Longest wavelength required for photoemission is 5396.74 A


## Example 14.19, Page 14.27¶

In [27]:

# Given
w = 2 # work function of sodium in eV
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C

#Calculations
lamda = ((h * c) / w) * (1 / e)

#Result
print "Threshold wavelength for photo emission is %d A"%(lamda * 1e10)


Threshold wavelength for photo emission is 6206 A


## Example 14.20, Page 14.27¶

In [35]:

# Given
k = 4 # maximum kinetic energy of electron in eV
w = 2.2 # work function of sodium in eV
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C

#Calculations
lambda_ = ((h * c) / (w * e))
lamda = (1 / ((((k * e)  / (h * c))) + (1 / lambda_)))

#Result
print "Threshold wavelength is %d A\nIncident electromagnetic wavelength is %.f A"%(lambda_ * 1e10,lamda * 1e10)


Threshold wavelength is 5642 A
Incident electromagnetic wavelength is 2002 A


## Example 14.21, Page 14.28¶

In [36]:

# Given
lamda = 3.5e-7 # wavelength of light in meter
i = 1 # intensity in W/m^2
p = 0.5 # percent of incident photon produce electron
a = 1 # surface area of potassium in cm^2
w = 2.1 # work function of potassium in eV
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C
m = 9.1e-31 # mass of an electron in kg

#Calculations
E = (((h * c) / lamda) * (1 / e) - w) * e
E_ = (p * a * 1e-4) / 100 # in W/cm^2
n = E_ / E

#Result
print "Maximum kinetic energy is %.3e J\nNumber of electrons emitted per sec from 1cm^2 area is %.2e"%(E,n)


Maximum kinetic energy is 2.314e-19 J
Number of electrons emitted per sec from 1cm^2 area is 2.16e+12


## Example 14.22, Page 14.28¶

In [37]:

# Given
lamda = 5.896e-7 # wavelength of first light in meter
lambda_ = 2.83e-7 # wavelength of second light in meter
V1 = 0.12 # stopping potential for emitted electrons for first light in V
V2 = 2.2 # stopping potential for emitted electrons for second light in V
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C

#Calculations
h = (e * (V2 - V1) / c) / (1 / lambda_ - 1 / lamda)

#Result
print "Value of Planck constant is %.2e J-sec"%h


Value of Planck constant is 6.04e-34 J-sec


## Example 14.23, Page 14.29¶

In [39]:
from math import *

# Given
lamda = 1e-10 # wavelength of light in meter
theta = 90 # angle at which scattered radiation is viewed in degree
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C
m = 9.1e-31 # mass of an electron in kg

#Calculations
delta_lambda = (h * (1 - cos(theta*pi/180))) / (m * c)

#Result
print "Compton shift is %.3f A"%(delta_lambda * 1e10)


Compton shift is 0.024 A


## Example 14.24, Page 14.29¶

In [41]:
from math import *
# Given
lamda = 1e-10 # wavelength of light in meter
theta = 90 # angle in degree
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C
m = 9.1e-31 # mass of an electron in kg

#Calculations
delta_lambda = (h * (1 - cos(theta*pi/180))) / (m * c)
E = (h * c) / delta_lambda

#Result
print "Compton shift is %.3f A\nEnergy of incident beam is %.3f MeV"%(delta_lambda * 1e10,E / 1.6e-13)


Compton shift is 0.024 A
Energy of incident beam is 0.512 MeV


## Example 14.25, Page 14.30¶

In [42]:
from math import *

# Given
E = 4 # enrgy of recoil electron in KeV
theta = 180 # scattered angle of photon in degree
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C
m = 9.1e-31 # mass of an electron in kg

#Calculations
p = sqrt(2 * E * 10**3 * e * m)
lamda = (2 * h * c) / (p * c + E * 10**3 * e)

#Result
print "Wavelength of incident beam is %.3f A"%(lamda * 1e10)


Wavelength of incident beam is 0.365 A


## Example 14.26, Page 14.31¶

In [44]:
from math import *

# Given
lamda = 1e-10 # wavelength of light in meter
theta = 90 # angle in degree
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C
m = 9.1e-31 # mass of an electron in kg

#Calculations
delta_lambda = (h * (1 - cos(theta*pi/180))) / (m * c)
E = (h * c) * ((1 / lamda) - (1 / (lamda + delta_lambda)))

#Result
print "Compton shift is %.3e m\nKinetic energy is %.f eV"%(delta_lambda,E / 1.6e-19)


Compton shift is 2.425e-12 m
Kinetic energy is 294 eV


## Example 14.27, Page 14.31¶

In [5]:
from math import *

# Given
lamda = 0.144e-10 # wavelength of x-ray in meter
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C
m = 9.1e-31 # mass of an electron in kg

#Calculations
theta = 180 # for maximum shift
d_lamda = (h * (1 - cos(theta*pi/180))) / (m * c)
E = (h * c) * ((1. / lamda) - (1. / (d_lamda+lamda)))

#Result
print "Maximum Compton shift is %.4f A\nKinetic energy is %.2f KeV"%(delta_lambda * 1e10,E / 1.6e-16)


Maximum Compton shift is 0.0485 A
Kinetic energy is 21.72 KeV


## Example 14.28, Page 14.32¶

In [7]:
from math import *

# Given
lamda = 0.2e-10 # wavelength of x-ray in meter
theta = 45 # scattered angle in degree
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C
m = 9.1e-31 # mass of an electron in kg

#Calculations
delta_lambda = (h * (1 - cos(theta*pi/180))) / (m * c)
E = (h * c) * ((1 / lamda) - (1 / (lamda + delta_lambda)))
theta_ = 180 # for maximum
delta_lambda_ = (h * (1 - cos(theta_*pi/180))) / (m * c)
lambda_ = lamda + delta_lambda_
E_k = h*c*(1/lamda - 1/lambda_)

#Result
print "Wavelength of x-ray is %.4f A\nMaximum kinetic energy %.2e J"%(lambda_ * 1e10,E_k)


Wavelength of x-ray is 0.2485 A
Maximum kinetic energy 1.94e-15 J


## Example 14.29, Page 14.33¶

In [8]:

# Given
h = 6.62e-34 # Planck constant in J-sec
v = 96 # speed of automobile in km/hr
e = 1.6e-19 # charge on an electron in C
m = 2e3 # mass of automobile in kg

#Calculations
v_ = v * (5. / 18)
lamda = h / (m * v_)

#Result
print "de-Broglie wavelength is %.2e m"%lamda


de-Broglie wavelength is 1.24e-38 m


## Example 14.30, Page 14.33¶

In [9]:
from math import *

# Given
v = 50 # potential differece in volt
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
e = 1.6e-19 # charge on an electron in C
m = 9.1e-31 # mass of an electron in kg

#Calculations
lamda =  h  / sqrt(2 * m * v * e)

#Result
print "de-Broglie wavelength is %.2f A"%(lamda * 1e10)


de-Broglie wavelength is 1.73 A


## Example 14.31, Page 14.33¶

In [10]:
from math import *

# Given
t = 300 # temperature in K
k = 1.37e-23 # Boltzmann's constant in J/K
h = 6.62e-34 # Planck constant in J-sec
e = 1.6e-19 # charge on an electron in C
m = 1.67e-27 # mass of neutron in kg

#Calculations
lamda = h / sqrt(3 * m * k * t)

#Result
print "Wavelength of thermal neutron is %.3f A"%(lamda * 1e10)


Wavelength of thermal neutron is 1.459 A


## Example 14.32, Page 14.34¶

In [11]:

# Given
v = 2e8 # speed of proton in m/sec
h = 6.62e-34 # Planck constant in J-sec
e = 1.6e-19 # charge on an electron in C
m = 1.67e-27 # mass of proton in kg

#Calculations
lamda = h / (m * v)

#Result
print "Wavelength of matter wave associated with proton is %.2e m"%lamda


Wavelength of matter wave associated with proton is 1.98e-15 m


## Example 14.33, Page 14.34¶

In [12]:

# Given
lamda = 0.1e-10 # DE Broglie wavelength associated with electron in M
h = 6.62e-34 # Planck constant in J-sec
e = 1.6e-19 # charge on an electron in C
m = 9.1e-31 # mass of electron in kg

#Calculations
V = h**2 / (2 * m* e * lamda**2)

#Result
print "Potential difference is %.2f KV"%(V * 10**-3)

Potential difference is 15.05 KV


## Example 14.34, Page 14.34¶

In [13]:
from math import *

# Given
v = 200 # potential differece in volt
h = 6.62e-34 # Planck constant in J-sec
c = 3e8 # speed of light in m/sec
q = 3.2e-19 # charge on an alpha particle in C
m = 4 * 1.67e-27 # mass of alpha particle in kg

#Calculations
lamda =  h  / sqrt(2 * m * v * q)

#Result
print "de-Broglie wavelength = %.2e m"%lamda


de-Broglie wavelength = 7.16e-13 m


## Example 14.35, Page 14.34¶

In [14]:
from math import *

# Given
t = 400 # temperature in K
k = 1.38e-23 # Boltzmann's constant in J/K
h = 6.62e-34 # Planck constant in J-sec
e = 1.6e-19 # charge on an electron in C
m = 4 * 1.67e-27 # mass of helium atom in kg

#Calculations
lamda = h / sqrt(3 * m * k * t)

#Result
print "de-Broglie wavelength = %.4f A"%(lamda * 1e10)


de-Broglie wavelength = 0.6294 A


## Example 14.36, Page 14.35¶

In [15]:

# Given
v = 2000 # velocity of neutron in m/sec
h = 6.62e-34 # Planck constant in J-sec
e = 1.6e-19 # charge on an electron in C
m = 1.67e-27 # mass of neutron in kg

#Calculations
lamda = h / (m * v)

#Result
print "de-Broglie wavelength is %.2f A"%(lamda * 1e10)


de-Broglie wavelength is 1.98 A


## Example 14.37, Page 14.35¶

In [18]:

# Given
lamda = 1e-10 # wavelength in m
h = 6.62e-34 # Planck constant in J-sec
e = 1.6e-19 # charge on an electron in C
m = 9.1e-31 # mass of electron in kg
m_ = 1.7e-27 # mass of neutron in kg

#Calculations
v = h / (m_ * lamda)
E = h**2 / (2 * m * lamda**2)
E_ = h**2 / (2 * m_ * lamda**2)

#Result
print "Energy for electron is %.f eV\nEnergy for neutron is %.3f eV"%(E / e,E_ / e)


Energy for electron is 150 eV
Energy for neutron is 0.081 eV


## Example 14.38, Page 14.36¶

In [20]:
from math import *

# Given
E1 = 500 # kinetic energy of electron in first case in eV
E2 = 50 # kinetic energy of electron in second case in eV
E3 = 1 # kinetic energy of electron in third case in eV
h = 6.62e-34 # Planck constant in J-sec
e = 1.6e-19 # charge on an electron in C
m = 9.1e-31 # mass of electron in kg

#Calculations
lambda1 = h / sqrt(2 * m * E1 * e)
lambda2 = h / sqrt(2 * m * E2 * e)
lambda3 = h / sqrt(2 * m * E3 * e)

#Result
print "de-Broglie wavelength of electron - \n(1) In first case is %.4f A \n(2) In second case is %.3f A \n(3) In third is %.3f A"%(lambda1*1e10,lambda2*1e10,lambda3*1e10)


de-Broglie wavelength of electron -
(1) In first case is 0.5486 A
(2) In second case is 1.735 A
(3) In third is 12.268 A


## Example 14.39, Page 14.36¶

In [22]:
from math import *

# Given
E1 = 1 # kinetic energy of neutron in first case in eV
E2 = 510 # kinetic energy of neutron in second case in eV
h = 6.62e-34 # Planck constant in J-sec
e = 1.6e-19 # charge on an electron in C
m = 1.67e-27 # mass of neutron in kg

#Calculations
lambda1 = h / sqrt(2 * m * E1 * e)
lambda2 = h / sqrt(2 * m * E2 * e)
r = lambda1 / lambda2

#Result
print "Ratio of de-Broglie wavelengths is %.2f:1"%r


Ratio of de-Broglie wavelengths is 22.58:1


## Example 14.40, Page 14.37¶

In [24]:
from math import *

# Given
E = 20 # kinetic energy of proton in MeV
E2 = 510 # kinetic energy of neutron in second case in eV
h = 6.62e-34 # Planck constant in J-sec
e = 1.6e-19 # charge on an electron in C
m = 1.67e-27 # mass of proton in kg
m_ = 9.1e-31 # mass of electron in kg

#Calculations
lambda1 = h / sqrt(2 * m * 10**6 * E * e)
lambda2 = h / sqrt(2 * m_ * E * 10**6 * e)
r = lambda2 / lambda1

#Result
print "Ratio of de-Broglie wavelengths is 1:%.f"%r


Ratio of de-Broglie wavelengths is 1:43


## Example 14.41, Page 14.37¶

In [26]:
from math import *

# Given
E = 1 # kinetic energy of proton in MeV
h = 6.62e-34 # Planck constant in J-sec
e = 1.6e-19 # charge on an electron in C
m = 1.67e-27 # mass of proton in kg

#Calculations
v = sqrt(2 * E * 1.6e-13 / m)

#Result
print "Velocity is %.2e m/sec"%v


Velocity is 1.38e+07 m/sec


## Example 14.42, Page 14.38¶

In [27]:

# Given
r = 1. / 20 # ratio of velocity of proton to the velocity of light
c = 3e8 # velocity of light in m/sec
h = 6.62e-34 # Planck constant in J-sec
e = 1.6e-19 # charge on an electron in C
m = 1.67e-27 # mass of proton in kg

#Calculations
v = r * c
lamda = h / (m * v)

#Result
print "de-Broglie wavelength is %.3e m"%lamda


de-Broglie wavelength is 2.643e-14 m


## Example 14.43, Page 14.38¶

In [37]:

# Given
lamda = 5.0e-7 # wavelength in m
c = 3.e8 # velocity of light in m/sec
h = 6.62e-34 # Planck constant in J-sec
e = 1.6e-19 # charge on an electron in C
m = 1.67e-27 # mass of proton in kg
m_ = 9.1e-31 # mass of electron in kg

#Calculations
E1 = h**2 / (2 * m * lamda**2)
E2 = h**2 / (2 * m_ * lamda**2)

#Results
print 'kinetic energy of proton(in J) =%.3e'%E1
print 'kinetic energy of electron(in J) =%.2e'%E2


kinetic energy of proton(in J) =5.248e-28
kinetic energy of electron(in J) =9.63e-25


## Example 14.44, Page 14.38¶

In [38]:
from math import *

# Given
n = 1 # no. of Bohr's orbit of hydrogen atom
c = 3e8 # velocity of light in m/sec
h = 6.62e-34 # Planck constant in J-sec
e = 1.6e-19 # charge on an electron in C
m = 9.1e-31 # mass of electron in kg

#Calculations
E = (13.6 / n**2) * e
lamda = h / sqrt(2 * m * E)

#Result
print "de-Broglie wavelength is %.1f A"%(lamda*1e10)


de-Broglie wavelength is 3.3 A


## Example 14.45, Page 14.38¶

In [39]:
from math import *

# Given
t = 300 # temperature in K
k = 1.376e-23 # Boltzmann's constant in J/K
c = 3e8 # velocity of light in m/sec
h = 6.62e-34 # Planck constant in J-sec
e = 1.6e-19 # charge on an electron in C
m_ = 4 * 1.67e-27 # mass of helium atom in kg
m = 1.67e-27 # mass of hydrogen atom in kg

#Calculations
lambda1 = h / sqrt(3 * m * k * t)
lambda2 = h / sqrt(3 * m_ * k * t)
r = lambda1 / lambda2

#Result
print "Ratio of de-Broglie wavelengths is %d:1"%r


Ratio of de-Broglie wavelengths is 2:1


## Example 14.47, Page 14.40¶

In [41]:
# Given
lamda = 1.2e-10 # DE Broglie wavelength in m
c = 3e8 # velocity of light in m/sec
h = 6.62e-34 # Planck constant in J-sec
e = 1.6e-19 # charge on an electron in C
m = 9.1e-31 # mass of electron in kg

#Calculations
v1 = h / (m * lamda)
v2 = h / (2 * m * lamda)

#Result
print "Group velocity is %.2e m/sec\nPhase velocity is %.2e m/sec"%(v1,v2)


Group velocity is 6.06e+06 m/sec
Phase velocity is 3.03e+06 m/sec