# Chapter 9: Quantum Physics¶

## Example 9.1, Page 279¶

In :
#Varaible declaration
e = 1.602e-19; # Charge of electron in Coloumb
lamda = 2e-10; # Wavelength of a photon in meters
h = 6.62e-34; # Planc's constant in Joule second
c = 3.e8; # Velocity og light in air in meter per second

#Calculations
E = (h*c)/(lamda*e);#Thermal conductivity of Ni
p = h/lamda;#The momentum of photon

#Results
print 'The energy of photon is ',round(E,3),'eV' #Incorrect answer in textbook
print 'The momentum of photon is ',p,'(kg.m)/s'

The energy of photon is  6198.502 eV
The momentum of photon is  3.31e-24 (kg.m)/s


## Example 9.2, Page 280¶

In :
#Varaible declaration
h = 6.62e-34; # Planck's constant J.s
v = 440e3; # Operating frequency of radio in Hertz
P = 20e3 ; # Power of radio transmitter in Watts

#Calculation
n = P/(h*v);# Let n be the number of photons emitted per second

#Result
print 'Number of photon emitted per second is ',round(n/1e30,2),'*10^30'

Number of photon emitted per second is  68.66 *10^30


## Example 9.3, Page 280¶

In :
#Varaible declaration
h = 6.62e-34; # Planck's constant in J.s
c = 3e8; # Velocity of ligth in air
t = 18000; # Time of glow - (5*3600) in seconds
P = 30 #Power in watts
lamda = 5893e-10; # Wavelength of emitted ligth in meters

#calculations
E = (h*c)/lamda; # Energy of a photon
n = (P*t)/E; # let n be the number of photons emitted in 5 hours

#Result
print 'Number of photons emitted in 5 hours is',round(n/1e24,3),'*10^24'

Number of photons emitted in 5 hours is 1.602 *10^24


## Example 9.4, Page 287¶

In :
from math import cos

#Varaible declaration
h = 6.62*1e-34; # Plancl's constant in J.s
c = 3*1e8; # Velocity of light in vacccum in m/s
m = 9.1*1e-31; # Mass of electron in Kg
l = 0.7078*1e-10 # Wavelength in meter
theta = 90;

#Calculations
delta = (h*(1-round(cos(theta)))/(m*c));
Nlambda = l + delta;

#Result
print 'The wavelength of scattered X-rays is %.4f A'%(Nlambda/1e-10)

The wavelength of scattered X-rays is 0.7320 A


## Example 9.5, Page 287¶

In :
from math import cos,degrees,radians

#Varaible declaration
m = 9.1e-31; # Mass of electron in kg
h = 6.62e-34; # Planck's constant in J.s
c = 3e8; # Velocity of light in vaccum
lamda = 1.8e18; # Frequency of the incident rays
theta = 180;#angle in degree

#Calculations
lamda = c/lamda;
Nlambda = lamda+delta;#'Wavelength of scattered X-rays

#Result
print 'Wavelength of scattered X-rays is %.4f A'%(Nlambda/1e-10)

Wavelength of scattered X-rays is 1.7152 A


## Example 9.6, Page 288¶

In :
from math import cos

#Varaible declaration
m = 9.1e-31; # Mass of electron in kg
h = 6.62e-34; # Planck's constant in Js
c = 3e8; # Velocity of light in vaccum
lamda = 1.12e-10; # Wavelength of light in meters
theta = 90;

#Calculations
delta = (h*(1-round(cos(theta))))/(m*c);
Nlambda = lamda + delta;#The wavelength of scattered X-rays
E = (h*c)*((1/lamda)-(1/Nlambda)) ;#Energy of electron

#Results
print 'The wavelength of scattered X-rays is %.3f A'%(Nlambda/1e-10)
print 'Energy of electron is %.2f *10^-17 J'%(E/1e-17)


The wavelength of scattered X-rays is 1.144 A
Energy of electron is 3.76 *10^-17 J


## Exampe 9.7, Page 289¶

In :
from math import cos,radians

#Varaible declaration
m = 9.1e-31; # Mass of electron in kg
h = 6.62e-34; # Planck's constant in Js
c = 3e8; # Velocity of light in vaccum
lamda = 0.03e-10; # Wavelength of light in meters
theta = 60;#angle in degrees

#Calculations
Nlambda = lamda + delta;
E = ((h*c)*((1./lamda)-(1./Nlambda)))/1.6e-19 ;#Energy of recoiling electron

#Result
print 'Energy of recoiling electron is %.3f MeV'%(E/1e+6)

Energy of recoiling electron is 0.119 MeV


## Eample 9.8, Page 289¶

In :
from math import cos,radians

#Varaible declaration
m = 9.1e-31; # Mass of electron in kg
h = 6.62e-34; # Planck's constant in Js
c = 3e8; # Velocity of light in vaccum
lamda = 0.5e-10; # Wavelength of light in meters
theta = 90;

#Calculations
Nlambda = lamda + delta;
E = (h*c)*((1./lamda)-(1./Nlambda)) ;

#Result
print 'Energy of electron is %.2f *10^-16 J'%(E/1e-16)

Energy of electron is 1.84 *10^-16 J


## Example 9.9, Page 290¶

In :
#Varaible declaration
m = 9.1e-31; # Mass of electron in kg
h = 6.62e-34; # Planck's constant in Js
c = 3e8; # Velocity of light in vaccum
lamda = 1.5e-10; # Wavelength of light in meters
E = 0.5e-16; # Energy of electron in J

#Calculation
Nlambda = ((h*c)/lamda)-E;#'Energy of scattered electron

#Result
print 'Energy of scattered electron is %.2f *10^-16 J'%(Nlambda/1e-16)

Energy of scattered electron is 12.74 *10^-16 J


## Example 9.10, Page 290¶

In :
from math import cos,radians

#Varaible declaration
lamda=0.022*1e-10;#wavelength in meters
th=45;#angle in degree
m=9.1*1e-31;
c=3*1e8;#velocity of light in free space
h=6.62*1e-34;#planck's constant

#Calculations&Results
x=cos(th);
print 'delta lemda is= %.3f A'%(dlamda/1e-10)
lamda1=lamda-dlamda;#wavelength of incident X-rays
print 'wavelength of incident X-rays %.3f A'%(lamda1/1e-10)

delta lemda is= 0.007 A
wavelength of incident X-rays 0.015 A


## Example 9.11, Page 314¶

In :
#Varaible declaration
a = 1e-10 # Width of box in meter
m = 9.1e-31; # Mass of electron in kg
h = 6.62e-34; # Planck's constant in Js
c = 3e8; # Velocity of light in vaccum
n = 1; # Single electron

#Calculation
E = (n**2 * h**2)/(8*m*a**2*1.6e-19);

#Result
print'Energy of electrons is %.1f n^2 eV'%E

Energy of electrons is 37.6 n^2 eV


## Example 9.12, Page 314¶

In :
#Varaible declaration
a = 1e-10 # Width of box in meter
m = 9.1e-31; # Mass of electron in kg
h = 6.62e-34; # Planck's constant in Js
c = 3e8; # Velocity of light in vaccum
n = 1; # Single electron

#Calculations
E = (h**2)/(8*m*a**2);#Energy of in lower level
p = h/(2*a);#Momentum

#Results
print 'Energy of in lower level %.f *10^-18 J'%(E/1e-18)
print'Momentum is %.2f *10^-24 (kg.m)/s'%(p/1e-24)

Energy of in lower level 6 *10^-18 J
Momentum is 3.31 *10^-24 (kg.m)/s


## Example 9.13, Page 315¶

In :
#Varaible declaration
a = 0.2e-9 # Width of box in meter
m = 9.1e-31; # Mass of electron in kg
h = 6.62e-34; # Planck's constant in Js
c = 3e8; # Velocity of light in vaccum
E5 = 230*1.6e-19 #  Energy of a particle in Volts in 5th antinode
n = 5;

#Calculations
E1 = E5/(n**2);
m = (h**2)/(8*E1*a**2);#Mass of electron

#Result
print 'Mass of electron is %.2f *10^-31 kg'%(m/1e-31)

Mass of electron is 9.30 *10^-31 kg


## Example 9.14, Page 316¶

In :
#Varaible declaration
n = 1; # Single particle
a = 50e-10; # Width of box in meter
deltax = 10e-10; # Intervel between particle

#Calculations
p = (2/a)*deltax;#The probability of finding the particle

#Result
print 'The probability of finding the particle is %.1f'%p

The probability of finding the particle is 0.4


## Example 9.15, Page 316¶

In :
from math import sqrt

#Varaible declaration
h = 6.62*1e-34; # Planck's constant
m = 1e-9; # Mass of particle in kg
t = 100; #Time reqired by the particle to cross 1 mm distance
a = 1e-3 ; # Width of box in m
v = 1e-5; # Velocity of particle in m/s

#Calculations
E = (0.5*m*v**2);
n = sqrt(8*m*a**2*E/(h**2));#The quantum state

#Result
print 'The quantum state is %.f*10^16 '%(n/1e+16)

The quantum state is 3*10^16


## Example 9.16, Page 317¶

In :
#Varaible declaration
h = 6.62e-34; # Planck's constant in J.s
m = 9.1e-31 # Mass of electron in kg
nk =1;
nl = 1;
nm = 1;
a = 0.5e-10 # Width of cubical box in meter

#Calculation
E = (h**2*(nk**2+nl**2+nm**2))/(8*m*a**2*1.6e-19);#The lowest energy level will have energy

#Result
print 'The lowest energy level will have energy %.f eV'%E

The lowest energy level will have energy 451 eV