The particle like properties of electromagnetic radiation

Example 3.1 Page 69

In [1]:
#initiation of variable
from math import sin,pi
w=0.250; theta=26.3;n=1 # n=1 for hydrogen atom and rest all are given values

#calculation
d=n*w/(2*sin(theta*pi/180));     # bragg's law

#result
print "Hence the atomic spacing in nm is",round(d,3);

Hence the atomic spacing in nm is 0.282

Example 3.2 Page 73

In [2]:
#initiation of variable
from math import pi,sin
I=120.0;r=0.1*10**-9;Eev=2.3   #I-intensity in W/m^2 r in m & E in electron volt
A=pi*r**2;K=1.6*10**-19;     # A=area and K is conversion factor from ev to joules

#calculation
t= Eev*K/(I*A);            #time interval

#result
print "The value of time interval was found out to be in sec is",round(t,3);

The value of time interval was found out to be in sec is 0.098

Example 3.3 Page 76

In [8]:
#initiation of variable
from math import pi,sin
w=650.0*10**-9;h=6.63*10**-34;c=3*10**8;  #given values and constant taken in comfortable units

#calculation
E=h*c/w; 
E1=E/(1.6*10**-19);

#result
print "The Energy of the electron in J ",E,"which is equivalent to in eV is ", round(E1,3);
print "The momentum of electron is p=E/c i.e is ", round(E1,3);

#part b
E2=2.40;                          #given energy of photon.

#calculation
w2=h*c*10**9/(E2*1.6*10**9);      #converting the energy in to eV and nm 

#result
print "The wavelength of the photon in m is",round(w2*10**28,0)

The Energy of the electron in J  3.06e-19 which is equivalent to in eV is  1.912
The momentum of electron is p=E/c i.e is  1.912
The wavelength of the photon in m is 518.0

Example 3.4 Page 78

In [5]:
#initiation of variable
hc=1240.0; phi=4.52               #both the values are in eV

#calcualtion
w1=hc/phi; 

#result
print "The cutoff wavelength of the tungsten metal in nm is ",round(w1,3);

#part b
w2=198.0;     #given value of wavelength 

#calculation
Kmax=(hc/w2)-phi;

#result
print 'The max value of kinetic energy in eV is',round(Kmax,3);

#part c
Vs=Kmax;

#result
print "The numerical value of the max kinetic energy is same as stopping potential in volts.Hence in V is",round(Vs,3);

The cutoff wavelength of the tungsten metal in fnm is  274.336
The max value of kinetic energy in eV is 1.743
The numerical value of the max kinetic energy is same as stopping potential in volts.Hence in V is 1.743

Example 3.5 Page 80

In [8]:
#initiation of variable
T1=293.0; Kw=2.898*10**-3;

#calculation
w1=Kw/T1;

#result
print "The wavelength at which emits maximum radiation in um. is",round(w1*10**6,3);

#part b
w2=650.0*10**-9; 
T2=Kw/w2;

#result
print 'The temperature of the object must be raised to in K. is',round(T2,3);

#part c
x=(T2/T1)**4; 

#result
print "Thus the thermal radiation at higher temperature in times the room (lower) tempertaure. is",round(x,3);

The wavelength at which emits maximum radiation in um. is 9.891
The temperature of the object must be raised to in K. is 4458.462
Thus the thermal radiation at higher temperature in times the room (lower) tempertaure. is 53612.939

Example 3.6 Page 82

In [9]:
#initiation of variable
#part a
from math import cos, sin, pi,atan
w1=0.24;wc=0.00243;theta=60.0;   #given values w=wavelength(lambeda)

#calculation
w2=w1+(wc*(1-cos(theta*pi/180))); 

#result
print "The wavelength of x-rays after scattering in nm is",round(w2,5);

#part b;
hc=1240;
E2=hc/w2;E1=hc/w1; 

#result
print "The energy of scattered x-rays in eV is",round(E2,3);

#part c
K= E1-E2; #The kinetic energy is the difference in the energy before and after the collision;

print "The kinetic energy of the x-rays in eV is",round(K,3);

#part d
phi2=atan(E2*sin(theta*pi/180)/(E1-E2*cos(theta*pi/180)))

#result
print "The direction of the scattered eletron in degrees is",round(phi2*180/pi,3);

The wavelength of x-rays after scattering in nm is 0.24121
The energy of scattered x-rays in eV is 5140.642
The kinetic energy of the x-rays in eV is 26.025
The direction of the scattered eletron in degrees is 59.749