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