Chapter 2 Particle nature of Radiation; The origin of Quantum theory¶

Example 2.2 Page no-12¶

In [27]:
#Given
E=40                                   #W
lembda=6000*10**-10                     #m
h=6.63*10**-34                          #Js
c=3*10**8                            #m/s

#Calculation
n=(E*lembda)/(h*c)

#Result
print"No. of photons emitted per second are given by ",round(n*10**-19,2),"*10**19"

No. of photons emitted per second are given by  12.07 *10**19


Example 2.3 Page no-12¶

In [31]:
#Given
a=3.2                                    #ev
energy=3.8                               #ev
e=1.6*10**-19

#Calculation
c=energy-a
Energy=c*e

#Result
print"Kinetic energy of the photoelectron is given by ",Energy,"Joule"

Kinetic energy of the photoelectron is given by  9.6e-20 Joule


Example 2.4 Page no-12¶

In [193]:
#Given
W=3.45                                     #ev
h=6.63*10**-34                              #Js
c=3*10**8                                   #m/s
e=1.6*10**-19

#Calculation
lembda=(h*c)/(W*e)

#Result
print"Maximum wavelength of photon is ",round(lembda*10**10,0),"A"

Maximum wavelength of photon is  3603.0 A


Example 2.5 Page no-12¶

In [197]:
#Given
W=3                                      #ev
h=6.63*10**-34
e=1.6*10**-19
lembda=3.0*10**-7                     #m
c=3*10**8                             #m/s

#Calculation
v0=(W*e)/h
v=c/lembda
E=h*(v-v0)
E1=(h*(v-v0))/(1.6*10**-19)
V0=E/e

#Result
print"(a) Threshold frequency ",round(v0*10**-15,2),"*10**15 HZ"
print"(b) Maximum energy of photoelectron ",round(E1,2),"eV"
print"(c) Stopping potential ",round(V0,2),"V"

(a) Threshold frequency  0.72 *10**15 HZ
(b) Maximum energy of photoelectron  1.14 eV
(c) Stopping potential  1.14 V


Example 2.6 Page no-13¶

In [88]:
#Given
v0=6*10**14                              #s**-1
h=6.63*10**-34
e=1.6*10**-19
V0=3

#Calculaton
W=h*v0
W0=(h*v0)/e
V=(e*V0+h*v0)/h

#Result
print"work function is given by ",round(W0,3),"ev"
print"frequency is given by ",round(V*10**-15,2),"*10**15 s-1"

work function is given by  2.486 ev
frequency is given by  1.32 *10**15 s-1


Example 2.7 Page no 13¶

In [200]:
#Given
lembda=6800.0*10**-10                             #m
h=6.6*10**-34
W=2.3                                          #ev
c=3*10**8                                      #m/s

#Calculation
E=((h*c)/lembda)/1.6*10**-19

#Result
print"Energy is ",round(E*10**38,2),"ev"
print"since the energy of incident photon is less then the work function of Na, photoelecrticemession is not possible with the given light."

Energy is  1.82 ev
since the energy of incident photon is less then the work function of Na, photoelecrticemession is not possible with the given light.


Example 2.8 Page no 14¶

In [201]:
#Given
lembda=3500*10**-10                               #m
h=6.6*10**-34
c=3*10**8                                         #m/s

#calculation
E=((h*c)/lembda)/1.6*10**-19

#Result
print"Energy is " ,round(E*10**38,2),"ev"
print"1.9 ev < E < 4.2 ev,only metal B will yield photoelectrons"

Energy is  3.54 ev
1.9 ev < E < 4.2 ev,only metal B will yield photoelectrons


Example 2.9 Page no 14¶

In [112]:
#Given
lembda=6.2*10**-6
W=0.1                                          #ev
h=6.6*10**-34                                  #Js
c=3*10**8                                      #m/s
e=1.6*10**-19

#Calculation
E=((h*c)/(lembda*e))-W

#Result
print"Maximum kinetic energy of photoelectron ",round(E,1),"ev"

Maximum kinetic energy of photoelectron  0.1 ev


Example 2.10 Page no 14¶

In [114]:
#given
e=1.60*10**-19                                  #C
slope=4.12*10**-15                              #Vs

#Calculation
h=slope*e

#Result
print"Value of plank's constant ",h,"Js"

Value of plank's constant  6.592e-34 Js


Example 2.11 Page no 15¶

In [118]:
#Given
W=2.26*1.6*10**-19                             #ev
v=10**6                                        #m/s
m=9*10**-31

#Calculation
V=((1/2.0)*m*v**2+W)/h

#Result
print"frequency of incident radiation ",round(V*10**-15,2),"*10**15 HZ"

frequency of incident radiation  1.23 *10**15 HZ


Example 2.12 Page no 15¶

In [202]:
#given
V1=.82                                            #volts
V2=1.85                                           #volts
lembda1=4.0*10**-7                                #m
lembda2=3.0*10**-7
e=1.6*10**-19
c=3.0*10**8                                     #m/s

#Calculation
lembda=(1/lembda2)-(1/lembda1)
h=(e*(V2-V1))/(c*lembda)

#Result
print"(a) plank's constant ",h,"Js"
print"(b) no, because the stopping potentialdepends only on the wavelength of light and not on its intensity."

(a) plank's constant  6.592e-34 Js
(b) no, because the stopping potentialdepends only on the wavelength of light and not on its intensity.


Example 2.13 Page no 16¶

In [131]:
#given
h=6.62*10**-34                              #Js
c=3*10**8                                   #m/s
lembda=4560.0*10**-10                         #m
p=1*10**-3                                  #W
a=0.5/100
e=1.6*10**-19

#calculation
E=(h*c)/lembda
N=p/E                                       #Number of photons incedent on the surface
n=N*a
I=n*e

#result
print"Photoelectric current ",round(I*10**6,2),"*10**-6 A"

Photoelectric current  1.84 *10**-6 A


Example 2.14 Page no 22¶

In [278]:
#given
m0=9.1*10**-31                            #Kg
c=3*10**8                                #m/s
h=6.6*10**-34                            #Js
v1=2.0*10**-10                           #m

#Calculation
import math
v= (h/(m0*c))*(1-(math.cos(90))*3.14/180.0)
v2=v+v1
v0=v2-v1
E=(h*c*(v0))/(v1*v2)
b=(1/(math.sin(90)*3.14/180.0))*((v2*10**-10/v1)-math.cos(90)*3.14/180.0)
angle=3.14/2.0-math.atan(b)

#Result
print "(a) the wavelength of scattered photon is ",round(v2*10**10,3),"A"
print"(b) The energy of recoil electron is ",round(E*10**17,2),"*10**-17 J"
print"(c) angle at which the recoil electron appears ",round(angle,2),"degree"

(a) the wavelength of scattered photon is  2.024 A
(b) The energy of recoil electron is  1.19 *10**-17 J
(c) angle at which the recoil electron appears  1.11 degree


Example 2.15 Page no 23¶

In [142]:
#Given
E=0.9                                #Mev
a=120                                #degree
m=9.1*10**-31                        #Kg
c=3*10**8                            #m/s

#calculation
b=((m*c**2)/1.6*10**-19)*10**32
energy=E/(1+2*(E/b)*(3/4.0))

#Result
print "energy of scattered photon ",round(energy,3),"Mev"

energy of scattered photon  0.247 Mev


Example 2.16 Page no 24¶

In [277]:
#Given
v1=2.000*10**-10                       #m
v2=2.048*10**-10                       #m
a=180                                  #degree
a1=60                                  #degree
h=6.6*10**-34
c=3*10**8

#Calculation
import math
b=(v2-v1)/(1-math.cos(a*3.14/180.0))
V=v1+b*(1-math.cos(60*3.14/180.0))
E=(h*c*(V-v1))/(V*v1)

#Result
print"(a) wavelength of radiation scattered at an angle of 60 degree ",round(V*10**10,3),"A"
print "(b) Energy of the recoiul electron is ",round(E*10**18,2),"*10**-18 J"

(a) wavelength of radiation scattered at an angle of 60 degree  2.012 A
(b) Energy of the recoiul electron is  5.9 *10**-18 J


Example 2.17 Page no 24¶

In [233]:
#Given
E=4*10**3*1.6*10**-19
m0=9.1*10**-31
b=6.4*10**-16
d=102.39*10**-16
h=6.3*10**-34
c=3*10**8

#Calculation
import math
p=math.sqrt(2*m0*E)
d=b+d
lembda=(2*h*c)/d

#Result
print"Wavelength of incident photon is ", round(lembda*10**10,2),"A"

Wavelength of incident photon is  0.35 A


Example 2.19 Page no 26¶

In [263]:
#Given
E=1.02                               #Mev
b=0.51

#Calculation
import math
alpha=E/b
a=1/(math.sqrt(2*(alpha+2)))
angle=2*(math.asin(a)*180/3.14)
e=E/(1.0+alpha*(1-(math.cos(angle*3.14/180.0))))

#Result
print"(a) Angle for symmetric scattering is ", round(angle,1),"degree"
print "(b) energy of the scattered photon is ",round(e,2),"Mev"

(a) Angle for symmetric scattering is  41.4 degree
(b) energy of the scattered photon is  0.68 Mev