Introduction To Special Relativity And Space Science (By S.P. Singh)

CHAPTER NUMBER 3 : DUAL NATURE OF LIGHT

EXAMPLE NUMBER 3.1 : (Page Number 135)

In [2]:
import math
   #Given that
h = 6.6e-34    # plank's constant
nu = 2e15    # frequency in Hz
phi1 = 6.72e-19
m = 9e-31
print "Standard formula Used   ( 1/2)*m*v**2 = h*nu - phi1"
v = math.sqrt((h * nu)/ m )    #calculation of maximum velocity of photoelectron
print "  Maximum velocity of photoelectron can be ",v,"  m/s.. "
Standard formula Used   ( 1/2)*m*v**2 = h*nu - phi1
  Maximum velocity of photoelectron can be  1211060.14164   m/s.. 

EXAMPLE NUMBER 3.2 : (Page Number 135)

In [3]:
import math
   #Given that
h = 6.6e-34    # plank's constant
lambda1_threshold = 2.4e-7    # threshold wavelength in cm
lambda1 = 2e-7    # wavelength of irradicated light in photo emmission
c = 3e8
print "  Standard formula Used   E = h * (nu1 – nu2)"
E = h * c * ((lambda1_threshold - lambda1)/(lambda1 *lambda1_threshold))/1.6e-19    # calculation of nergy of photoelectrons
print "  Energy of photoelectrons emitted is ",E,"  eV"
  Standard formula Used   E = h * (nu1 – nu2)
  Energy of photoelectrons emitted is  1.03125   eV

EXAMPLE NUMBER 3.3 : (Page Number 136)

In [4]:
import math
   #Given that
applied_voltage = 4e4    # in volt
h = 6.624e-34    # plank's constant
c = 3e8    # speed of light
e = 1.6e-19    # charge on electron
print "  Standard formula Used   E = h*c/lambda1"
lambda1 = h * c / ( e * applied_voltage) *1e10    #calculation of Shortest wavelength emitted
print "  Shortest wavelength emitted is ",lambda1," Angstrom." 
  Standard formula Used   E = h*c/lambda1
  Shortest wavelength emitted is  0.3105  Angstrom.

EXAMPLE NUMBER 3.4 : (Page Number 136)

In [5]:
import math
   #Given that
E = 1e3    # energy of moving electron in eV
h = 6.624e-34    # plank's constant
c = 3e8    # speed of light
e = 1.6e-19    # charge on electron
m_e = 9.1e-31
print "  Standard formula Used   E =(1/2)*m *v**2"
v = math.sqrt(2 * E * 1.6e-19/ m_e)    #calculation of Velocity of moving electron
print "  Velocity of moving electron is ",v,"  m/s."
  Standard formula Used   E =(1/2)*m *v**2
  Velocity of moving electron is  18752289.2375   m/s.

EXAMPLE NUMBER 3.5 : (Page Number 137)

In [6]:
import math
   #Given that
phi1 = 6    # work function in eV
h = 6.624e-34    # plank's constant
c = 3e8    # speed of light
e = 1.6e-19    # charge on electron
m_e = 9.1e-31
print "  Standard formula Used   phi1 = h * nu"
lambda1 = h * c / (phi1 * e) * 1e10   #calculation of Longest wavelength to eject electron
print "  Longest wavelength to eject electron is ",lambda1,"  Angstroms. "
  Standard formula Used   phi1 = h * nu
  Longest wavelength to eject electron is  2070.0   Angstroms. 

EXAMPLE NUMBER 3.6 : (Page Number 137)

In [12]:
import math
   #Given that
pi = 3.14
theta1 =pi/2    # scattering angle of photon
h = 6.624e-34    # plank's constant
c = 3e8    # speed of light
e = 1.6e-19    # charge on electron in coloumb
m_e = 9.1e-31    # mass of electron in kg
print "  Standard formula Used   delta_lambda1 = h * (1 - cos (theta1 )) / ( m_e * c)"
delta_lambda1 = h * (1 - math.cos(theta1 )) /( m_e * c)    #calculation of Change in wavelength of electron
print "  Change in wavelength of electron is ",round(delta_lambda1*1e10,4),"  Angstrom. " 
  Standard formula Used   delta_lambda1 = h * (1 - cos (theta1 )) / ( m_e * c)
  Change in wavelength of electron is  0.0242   Angstrom. 

EXAMPLE NUMBER 3.7 : (Page Number 137)

In [13]:
import math
   #Given that
angle = pi/2    # scattering angle of photon
h = 6.624e-34    # plank's constant
v = 2e6    # speed of particle
e = 1.6e-19    # charge on electron
m = 1e-3    # mass of particle in kg
print "  Standard formula Used   lambda1 = h / (m * v)"
lambda1 = h / (m * v)    #calculation of de Broglie wavelength of particle
print "  de Broglie wavelength of particle is ",lambda1,"  m." 
print "  Here the de Broglie wavelength is too small to be detected. This wavelength is far smaller than the wavelength of X ray.  Hence diffraction experiment with such a stream of particle will not be successful."
  Standard formula Used   lambda1 = h / (m * v)
  de Broglie wavelength of particle is  3.312e-37   m.
  Here the de Broglie wavelength is too small to be detected. This wavelength is far smaller than the wavelength of X ray.  Hence diffraction experiment with such a stream of particle will not be successful.

EXAMPLE NUMBER 3.8a : (Page Number 138)

In [21]:
import math
   #Given that
lambda1 = 4.3e-7    # wavelength of light in meter
phi1_Ni = 5    # work function of nickel in eV
h = 6.624e-34    # plank's constant
c = 3e8    # speed of light
m_e = 9.1e-31    # mass of electron in kg
lambda1_threshold = h * c / (phi1_Ni*1e-19)    #calculation of longest wavelength required
if lambda1_threshold < lambda1:
    print "  As the threshold wavelength is less than wavelength of incident radiation   So electron will not be ejected  "
else:
    v = math.sqrt((2*h*c*(lambda1-lambda1_threshold))/(m*lambda1_threshold*lambda1))    #calculation of ejected velocity Electron
    print "  As the threshold wavelength is greater than wavelength of incident radiation So electron will be ejected with velocity ",v,". "
  As the threshold wavelength is less than wavelength of incident radiation   So electron will not be ejected  

EXAMPLE NUMBER 3.8b : (Page Number 138)

In [24]:
import math
   #Given that
lambda1 = 4.3e-7    # wavelength of light in meter
phi1_K = 2.3    # work function of nickel in eV
h = 6.624e-34    # plank's constant
c = 3e8    # speed of light
m_e = 9.1e-31    # mass of electron in kg
lambda1_threshold = h * c / (phi1_K *1.6e-19)    #calculation of longest wavelength required
if lambda1_threshold < lambda1:
    print "As the threshold wavelength is less than wavelength of incident radiation Solectron will not be ejected  "
else:
    v = math.sqrt((2* h * c *( lambda1_threshold - lambda1)) / (m_e * lambda1_threshold * lambda1 ))    #calculation of ejected velocity Electron
    print "  As the threshold wavelength is greater than wavelength of incident radiation So   electron will be ejected with velocity ",v," m/s. "
  As the threshold wavelength is greater than wavelength of incident radiation So   electron will be ejected with velocity  454862.700897  m/s. 

EXAMPLE NUMBER 3.9 : (Page Number 139)

In [27]:
import math
   #Given that
d = 3.04    # inter layer separation in Angstrom
theta1 = 14.7    # in degree
n = 2    # order of brags reflection
print "  Standard formula Used   2 * d * sin(theta1) = n * lambda1"
lambda1 = 2 * d * math.sin( theta1 * (pi /180))/ n      #calculation of wavelength making second order Braggs reflection
print "   Second order brags reflection occurs at ",theta1," degree for the wavelength ",lambda1," Angstrom "
  Standard formula Used   2 * d * sin(theta1) = n * lambda1
   Second order brags reflection occurs at  14.7  degree for the wavelength  0.771041684638  Angstrom 

EXAMPLE NUMBER 3.10a : (Page Number 139)

In [29]:
import math
   #Given that
lambda1 = 0.52    # wavelength in angstrom
theta1 = 5    # in degree
n = 1    # order of brags reflection
print "  Standard formula Used   2 * d * sin(theta1) = n * lambda1 "
d = n * lambda1 / (2 * math.sin (theta1 * pi / 180))
   #calculation of separation between adjacent layers of crystals
print "  Separation between adjacent layers of crystals is ",d," angstrom. "
  Standard formula Used   2 * d * sin(theta1) = n * lambda1 
  Separation between adjacent layers of crystals is  2.98467470674  angstrom. 

EXAMPLE NUMBER 3.10b : (Page Number 139)

In [32]:
import math
   #Given that
n = 2    # order 
lambda1 = 5.2e-11    # wavelength in Angstrom 
d = 2.98e-10    # interatomic separation in Angstrom
print "  Standard formula Used   2 * d * sin(theta1) = n * lambda1 "
theta1_rad = math.asin ( (n * lambda1) / ( 2 * d))    #calculation of angle for secondary maxima in radian
theta1_deg = theta1_rad * 180 / pi    #calculation of angle for secondary maxima in degree
print "  Angle for secondary maxima is  ",theta1_deg,". "
  Standard formula Used   2 * d * sin(theta1) = n * lambda1 
  Angle for secondary maxima is   10.0544644405 . 

EXAMPLE NUMBER 3.11 : (Page Number 140)

In [33]:
import math
   #Given that
nu = 3.2e19    # frequency in hartz
theta1 = 90    # angle of scattered photon in degree
m_e = 9.1e-31    # mass of electron in Kg
c = 3e8    # speed of light in m/s
h = 6.626e-34    # plank's constant
print "  Standard formula Used   delta_lambda1 = h * (1 - math.cos (theta1 )) / ( m_e * c)"
lambda1 = c / nu   #calculation of incident wavelength
lambda1_shift = h *(1 - math.cos(theta1 * pi / 180))/ ( m_e * c)    #calculation of shift in wavelength
lambda11 = lambda1 + lambda1_shift   #calculation of wavelength of scattered photon
nu1 = c / lambda11   #calculation of Frequency after scattering  
print "  Frequency after scattering  is ",nu1," Hz. "
  Standard formula Used   delta_lambda1 = h * (1 - math.cos (theta1 )) / ( m_e * c)
  Frequency after scattering  is  2.54233550954e+19  Hz. 

EXAMPLE NUMBER 3.12 : (Page Number 140)

In [34]:
import math
   #Given that
r = 1e-14    # radius of nucleus of atom in meter
h = 6.626e-34    # Plank's constant
print "  Standard formula Used   delta_p * delta_x >= h /(2*pi)"
del_x = 2 * r    #calculation of Uncertainty in position
del_p = h / (2 * pi * del_x)    #calculation of Uncertainty in momentum
print "  Uncertainty in momentum is ",del_p," Kg-m/s. "
  Standard formula Used   delta_p * delta_x >= h /(2*pi)
  Uncertainty in momentum is  5.27547770701e-21  Kg-m/s. 

EXAMPLE NUMBER 3.13 : (Page Number 140)

In [35]:
import math
   #Given that
v = 300    # speed of electron in m/s
accuracy = 1e-4    # accuracy in speed
h = 6.6e-34    # Plank's constant
m_e = 9.1e-31    # mass of electron in Kg
print "  Standard formula Used   delta_p * delta_x >= h /(2*pi)"
del_p =  accuracy * m_e * v     #calculation of Uncertainty in  momentum
del_x = h / (4 * pi * del_p)     #calculation of Uncertainty in position
print "  Uncertainty in position of electron is ",del_x*1000," mm. "
  Standard formula Used   delta_p * delta_x >= h /(2*pi)
  Uncertainty in position of electron is  1.92482676559  mm. 

EXAMPLE NUMBER 3.14 : (Page Number 141)

In [36]:
import math
   #Given that
lambda11 = 6560    # wavelength in Angstrom
n1 = 1    # transition state no
n2 = 2    # transition state no
n3 = 3    # transition state no.
print "  Standard formula Used    For Balmer Series   1/lambda1 =  R*(1-(1/n)**2)    For Lyman series   1/lambda1 =  R*((1/2)**2 -(1/n)**2)"
lambda12 = (n2**2 * n1**2) *(n3**2 - n2**2) /( (n2**2 - n1**2) * (n3**2 * n2**2)) * lambda11     #calculation of Wavelength of first line of Lyman series
print "   Wavelength of first line of Lyman series is ",lambda12," Angstrom. "
  Standard formula Used    For Balmer Series   1/lambda1 =  R*(1-(1/n)**2)    For Lyman series   1/lambda1 =  R*((1/2)**2 -(1/n)**2)
   Wavelength of first line of Lyman series is  0  Angstrom. 

EXAMPLE NUMBER 3.15 : (Page Number 142)

In [39]:
import math
   #Given that
m = 2e-3    # mass of linear harmonic oscillator in kg
k = 100    # spring constant in N/m
h = 6.6e-34    # Plank's constant 
print "  Standard formula Used   f =  math.sqrt(k / m )   U = 1/2* h * nu "
nu1 = math.sqrt(k / m ) / (2 * pi)    #calculation of frequency of linear harmonic oscillator
U = 1/2* h * nu1    #calculation of Zero point energy of a linear harmonic oscillator  
print "  Zero point energy of a linear harmonic oscillator  is ",U," J."
  Standard formula Used   f =  math.sqrt(k / m )   U = 1/2* h * nu 
  Zero point energy of a linear harmonic oscillator  is  0.0  J.

EXAMPLE NUMBER 3.16a : (Page Number 142)

In [40]:
import math
   #Given that
R = 1.097    # Rydberg’s constant
n1 = 1    # transition state no
n2 = 2    # transition state no
print "  Standard formula Used   For Lyman series 1/lambda1 = R*((1/2) **2 - (1/n) **2)"
nu1 = R * (n2**2 - n1**2) / (n1**2 * n2**2)    #calculation of frequency of first line of Lyman series
lambda11 = 1/ nu1   #calculation of Wavelength of first line of Lyman series
print "  Wavelength of first line of Lyman series is ",lambda11 *1000," Angstrom. "
  Standard formula Used   For Lyman series 1/lambda1 = R*((1/2) **2 - (1/n) **2)
  Wavelength of first line of Lyman series is  1215.43603768  Angstrom. 

EXAMPLE NUMBER 3.16b : (Page Number 142)

In [41]:
import math
   #Given that
R = 1.097    # Rydberg’s constant
n1 = 1    # transition state no
n2 = 3    # transition state no
print "  Standard formula Used   For Lyman series 1/lambda1 =  R*((1/2)**2 -(1/n)**2)"
nu1 = R * (n2**2 - n1**2) / (n1**2 * n2**2)    #calculation of frequency of first line of Lyman series
lambda11 = 1/ nu1    #calculation of Wavelength of first line of Lyman series
print "  Wavelength of second line of Lyman series is ",lambda11 *1000 ," Angstrom. "
  Standard formula Used   For Lyman series 1/lambda1 =  R*((1/2)**2 -(1/n)**2)
  Wavelength of second line of Lyman series is  1025.52415679  Angstrom. 

EXAMPLE NUMBER 3.17 : (Page Number 143)

In [42]:
import math
   #Given that
lambda11 = 4700    # wavelength in Angstrom
lambda12 = 1.4e-5   #wavelength in cm 
temp1 = 6174    # temperature of a black of in kelvin
print "  Standard formula Used   lambda1 * T = constant"
temp2 = lambda11 * temp1 / (lambda12 * 1e8)    #calculation of temperature
print "  Blackbody will emit  wavelength 1.4e-5 cm at ",temp2," K."
  Standard formula Used   lambda1 * T = constant
  Blackbody will emit  wavelength 1.4e-5 cm at  20727.0  K.

EXAMPLE NUMBER 3.19a : (Page Number 144)

In [46]:
import math
   #Given that
lambda1 = 1    # wavelength in Angstrom
theta1 = 90    # angle of scattered photon in degree
m_e = 9.11e-31    # mass of electron in Kg
c = 3e8    # speed of light in m/s
h = 6.63e-34    # plank's constant
print "  Standard formula Used   delta_lambda1 = h * (1 - math.cos (theta1 )) / ( m_e * c)"
lambda1_shift = h *(1 - math.cos(theta1 * pi / 180))/ ( m_e * c)    #calculation of Change in frequency  
print "  Change in frequency  is ",round(lambda1_shift * 1e10,4)," Hz. "
  Standard formula Used   delta_lambda1 = h * (1 - math.cos (theta1 )) / ( m_e * c)
  Change in frequency  is  0.0242  Hz. 

EXAMPLE NUMBER 3.19b : (Page Number 144)

In [48]:
import math
   #Given that
lambda11 = 1    # wavelength in Angstrom
lambda12 = 1.0243    # wavelength in Angstrom
c = 3e8    # speed of light in m/s
h = 6.63e-34    # plank's constant
print "  Standard formula Used   E= h *(nu1 – nu2)"
K = h * c * (( lambda12 - lambda11 )/ (lambda11 * lambda12 )) *(10e9 / 1.6e-19)      #calculation of Kinetic energy imparted to recoiling
print "  Kinetic energy imparted to recoiling electron is ",round(K,4)," eV."
  Standard formula Used   E= h *(nu1 – nu2)
  Kinetic energy imparted to recoiling electron is  294.913  eV.

EXAMPLE NUMBER 3.20 : (Page Number 145)

In [50]:
import math
   #Given that
theta1 = 90    # angle of scattered photon in degree
E_rest = 938.3    # rest mass energy of a proton in MeV
E = 12   # energy of scattered proton in Mev
c = 3e8    # speed of light in m/s
h = 6.63e-34    # plank's constant
print "  Standard formula Used   delta_lambda1 = h * (1 - math.cos (theta1 )) / ( m_e * c)"
lambda1 = h * c / ( E * 1.6e-13)    #calculation of incident wavelength
lambda11 = lambda1 + h * c / (E_rest * 1.6e-13)       #calculation of wavelength of scattered photon
print "  wavelength of scattered photon is  ",round(lambda11 * 1e10,4)," Angstrom. "
  Standard formula Used   delta_lambda1 = h * (1 - math.cos (theta1 )) / ( m_e * c)
  wavelength of scattered photon is   0.001  Angstrom. 

EXAMPLE NUMBER 3.21 : (Page Number 146)

In [52]:
import math
   #Given that
lambda11 = 1.321    # wavelength of L- alpha line for platinum
lambda12 = 4.174    # wavelength of l - alpha line of unknown substance 
z1= 78   # atomic number of platinum
c = 3e8    # speed of light in m/s
b = 7.4    # constant  for L - alpha line
print "  Standard formula Used   math.sqrt(nu1)= a*(Z-b)"
z2 = b + (z1 - b) * math.sqrt(lambda11 / lambda12)    #calculation of the unknown substance has atomic number
print"  The unknown substance has atomic number ",round(z2,4),". "
  Standard formula Used   math.sqrt(nu1)= a*(Z-b)
  The unknown substance has atomic number  47.1173 . 

EXAMPLE NUMBER 3.22 : (Page Number 146)

In [59]:
import math
   #Given that
h = 6.6e-34    # plank's constant
m_e = 9.1e-31    # mass of electron in kg
L = 1e-10    # length of box of particle in m
print "  Standard formula Used   E= h**2 * (n_x**2+n_y**2+n_z**2) / (8*m*L**2)"
sum = 0 
n_y = 1
for n_x in range(1,3):
    for n_z in range(1,2):
                sum = n_x+n_y+n_z
                if sum<6:
                     E = h**2 * (n_x**2+n_y**2+n_z**2)/ (1.6e-19*8*m_e*L**2)    # calculation of energy
                print "    E",n_x,"",n_y,"",n_z," is  ",round(E,4)," eV. "
                
  Standard formula Used   E= h**2 * (n_x**2+n_y**2+n_z**2) / (8*m*L**2)
    E 1  1  1  is   112.1909  eV. 
    E 2  1  1  is   224.3819  eV.