# Chapter 2: Lasers¶

## Example 1, Page No: 2.40¶

In :
# Finding Number of electron hole pairs

import math

# Variable Declaration
A           = 4*10**-6;              #  Receiving area of photo detector
I           = 200;                   # Intensity in W/m^2
h           = 6.625*10**-34;          # planck's constant
c           = 3*10**8;                # vel. of light in m/s
lamda       = 0.4*10**-6;             # wavelength of light in m

#Calculations
v           = c/lamda;              # frequency
NOP         = I*A/(h*v)             # number of photons

#since each photon generates an electron hole pair, the number of photons is equal to number of electron hole pairs

# Result

print 'Number of electron hole pairs = %3.2e '%NOP;

Number of electron hole pairs = 1.61e+15


## Example 2, Page No:2.40¶

In :
# Finding Wavelength

import math

# Variable Declaration
Eg          = 2.8;                   # bandgap energy in eV
h           = 6.625*10**-34;          # plank's constant
c           = 3*10**8;                # vel. of light in m/s
q           = 1.602*10**-19;          # charge of electron

# Calculations
E           = Eg*q                 # eV to joules conversion
lamda       = h*c/E;               # wavelength

#Result
print 'wavelength = %3.1f' %(lamda*10**10), 'Å(Blue Colour)';

wavelength = 4430.8 Å(Blue Colour)


## Example 3, Page No:2.41¶

In :
# Finding Energy bandgap

import math;

# Variable Declaration
h           = 6.625*10**-34;          # plank's constant
c           = 3*10**8;                # vel. of light in m/s
lamda       = 1.55*10**-6;            # wavelength of light in m
q           = 1.6*10**-19;            # charge of electron

#Calculations
Eg          = (h*c)/lamda;           # band gap energy in joules
E           = Eg/q                   # bang gap energy in eV

# Result
print 'Energy bandgap Eg   = %3.4f'%E, 'eV';

Energy bandgap Eg   = 0.8014 eV


## Example 4, Page No:2.41¶

In :
# Finding Number of photons required to do one Joule of work

import math

# Variable Declaration
h           = 6.625*10**-34;         # plank's constant
c           = 3*10**8;               # vel. of light in m/s
lamda       = 4961*10**-10;          # wavelength of light in m

# Calculations
E           = (h*c)/lamda;           # energy in joules
N           = 1/E

# Result
print 'Number of photons required to do one Joule of work  = %3.4e'%N,'/m^3';

Number of photons required to do one Joule of work  = 2.4961e+18 /m^3


## Example 5, Page No:2.41¶

In :
# Finding Wavelength Limit
# import math

# Variable Declaration
E           = 0.02;                 # ionisation energy in eV
h           = 6.625*10**-34;         # plank's constant
c           = 3*10**8;               # vel. of light in m/s
q           = 1.6*10**-19;           # charge of electron

# Calculations

lamda       = h*c/(E*q)              # long wavelength limit in m

# Result

print 'long wavelength limit = %3.3e' %lamda,' m';

long wavelength limit = 6.211e-05  m


## Example 6, Page No:2.42¶

In :
# Finding Wavelength

import math;

# Variable Declaration
E           = 1.44;                 # Bandgap energy in eV
h           = 6.625*10**-34;        # plank's constant
c           = 3*10**8;              # vel. of light in m/s
q           = 1.6*10**-19;          # charge of electron

# Calculations

lamda       = h*c/(E*q)              # Wavelength of GaAs laser

# Result
print 'Wavelength of GaAs laser = %3.1f'%(lamda*10**10),' Å';

Wavelength of GaAs laser = 8626.3  Å


In :
# Finding Energy of the first excited state

import math;

# Variable Declaration
h           = 6.625*10**-34;         # planck's constant
c           = 3*10**8;                # vel. of light in m/s
lamda       = 5890*10**-10;           # wavelength of light in m
q           = 1.6*10**-19;            # charge of electron

# Calculations
Eg          = (h*c)/lamda;          # energy in joules
E           = Eg/q                  # energy in eV

# Result
print 'Energy of the first excited state  = %3.3f' %E,'eV';

Energy of the first excited state  = 2.109 eV


In :
# Finding The ratio between the stimulated emission and apontaneous emission

import math;

# Variable Declaration
h           = 6.625*10**-34;          # planck's constant
c           = 3*10**8;                # vel. of light in m/s
lamda       = 5890*10**-10;           # wavelength of light in m
k           = 1.38*10**-23;           # Boltzmann constant
Tc          = 280                     # Temperature in centigrades

# Calculations
T           = Tc+273;               # temperature in kelvin
R           = 1/((math.exp((h*c)/(k*T*lamda))) - 1);     # ratio of stimulated emission to spontaneous emission

# Result
print 'The ratio between the stimulated emission and apontaneous emission = %3.3e' %R;

The ratio between the stimulated emission and apontaneous emission = 6.264e-20


In :
# Finding The No. of Photons emitted per minute

import math;

# Variable Declaration
h           = 6.625*10**-34;          # planck's constant
c           = 3*10**8;                # vel. of light in m/s
lamda       = 6328*10**-10;           # wavelength of He-Ne laser source in m
q           = 1.6*10**-19;            # charge of electron
P           = 3*10**-3                # output power of the He-Ne source in watts or J/sec

# Calculations
v           = c/lamda               # frequency of the photon emitted by the laser beam
E           = h*v;                  # energy of a photon in joules
Po          = P*60;                 # conversion from J/sec to J/min
N           = Po/E;                 # No of photons emitted per minute

# Result
print 'The No. of Photons emitted per minute = %3.3e' %N, 'photons/minute';

The No. of Photons emitted per minute = 5.731e+17 photons/minute


In :
# Finding The No. of Photons emitted per hour

import math;

# Variable Declaration
h           = 6.625*10**-34;          # planck's constant
c           = 3*10**8;                # vel. of light in m/s
lamda       = 9.6*10**-6;             # wavelength of CO2 laser source in m
q           = 1.6*10**-19;            # charge of electron
P           = 10*10**3                # output power of the CO2 laser source in watts or J/sec

# Calculations
v           = c/lamda               # frequency of the photon emitted by the laser beam
E           = h*v;                  # energy of a photon in joules
Po          = P*60*60;              # conversion fro J/sec to J/hour
N           = Po/E;                 # No of photons emitted per hour

# Result
print 'The No. of Photons emitted per hour = %3.3e'%N,' photons/hour';

The No. of Photons emitted per hour = 1.739e+27  photons/hour


In :
import math;

# Variable Declaration
h           = 6.625*10**-34;          # planck's constant
c           = 3*10**8;                # vel. of light in m/s
lamda       = 10*10**-2;              # wavelength for microwave region in m
T           = 300                     # Temperature in Kelvin
Kb          = 1.38*10**-23            # Boltzmann constant

# Calculations
#  let R    = Rsp/Rst
R           = math.exp((h*c)/(lamda*Kb*T)) - 1 ;     # ratio of spontaneous to stimulated emission
if R<1:
print 'Since the spontaneous emission is lesser than stimulated emission \n hence MASER action is possible at thermal equilibrium'

Since the spontaneous emission is lesser than stimulated emission
hence MASER action is possible at thermal equilibrium


In :
import math;

# Variable Declaration
h           = 6.625*10**-34;          # planck's constant
c           = 3*10**8;                # vel. of light in m/s
lamda       = 5000*10*8-10;           # wavelength for optical region in m
T           = 300                     # Temperature in Kelvin
Kb          = 1.38*10**-23            # Boltzmann constant

# Calculations
# let R    = Rsp/Rst
R           = math.exp((h*c)/(lamda*Kb*T)) - 1;     # ratio of spontaneous to stimulated emission
if R<1:
print 'Since the spontaneous emission is lesser than stimulated emission \n hence LASER action is possible at thermal equilibrium'
else:
print 'Since the spontaneous emission is more predominant than stimulated emission\nhence LASER action is not possible at optical frequencies under thermal equilibrium';

Since the spontaneous emission is lesser than stimulated emission
hence LASER action is possible at thermal equilibrium


In :
import math

# Variable Declaration
h           = 6.625*10**-34;         # plank's constant
c           = 3*10**8;               # vel. of light in m/s
lamda       = 5511.11*10**-10;        # wavelength of green LED light in m
q           = 1.6*10**-19;            # charge of electron

# Calculations
Eg          = (h*c)/lamda;          # band gap energy in joules
E           = Eg/q                  # bang gap energy in eV

# Result
print ' Eg   = %3.2f' %E,'eV';

 Eg   = 2.25 eV