#Variable declaration
Ft= 660 #frequency of tuning fork, Hz
Fo= 5.00*(10**14) #frquency of atomic oscillator, Hz
Ef= 0.04 #vibrational energy of tuning fork, J
h= 6.63*(10**(-34)) #Planck's constant, J.s
#Calculation
E1= h*Ft #Total energy of tuning fork, J
E2= h*Fo #Total energy of atomic oscillator, J
E2= E2/(1.60*(10**(-19))) #converting to eV
#Result
print"(a).Energy of tuning fork is:%.3g"%E1,"J"
print"(b).Energy of atomic oscillator is:",round(E2,2),"eV(approx)"
#Variable declaration
l= 350 #Wavelength of UV light, nm
i= 1.00 #intensity of UV light, W/m**2
#Part (a)
l= l*10**(-9) #converting to m
Ep= (1.24*(10**(-6)))/l #energy of photon, using Eqn (2.11) on Page 66, e.V
t= 2.2 #work function of Potassium surface, eV
#Calculation
KEmax= Ep-t #Max KE of the phototelectrons, eV
#Part (b)
A= 1.00 #Surface area, cm**2
A= A* 10**(-4) #converting to m**2
E= 5.68*(10**(-19)) #Photon energy, J
Np= i*A/E #number of incident photon, per second
Ne= (0.0050)*Np #number of photoeectrons emitted, per second
#Result
print"(a).Maximum KE of photoelectrin is: ",round(KEmax,1),"eV"
print"(b).Rate of emission of photoelectrons is:%.2g"%Ne,"photoelectrons/s"
#Variable declaration
AP= 50000 #Accelerating potential of the x-ray machine, V
l= (1.24*(10**(-6)))/AP*(10**(9)) #Minimum wavelength, nm
#Calculation
Fmax= 3*(10**8)/(l*(10**(-9))) #Maximum frequency, Hz
#Result
print"Minimum wavelength possible is: ",l,"nm"
print"Maximum frequency possible is: %.3g"%Fmax,"Hz"
#Variable declaration
#part (a)
l= 10.0 #wavelength of x-ray, pm
r= 45.0 #angle of scattered articles, degree
lc= 2.426*(10**(-12)) #Compton wavelength for electron, m
#Calculation
import math
k= math.cos(math.radians(45))
lc= lc* 10.0**12 # converting to pm
print
l2= l+ lc*(1.0-k) #using Eqn 2.23
#Part (b)
lmax= l+(lc*2) #for (1-k)=2
#Part (c)
h= 6.63*(10**(-34)) #Planck's constant, J.s
c= 3*10**8 #velocity of light, m/s
c=c*10**12 #converting to pm/s
KEmax= (h*c)*((1/l)-(1/lmax)) #J
#Result
print"(a):The wavelength of scattered x-ray is: ",round(l2,1),"pm"
print"(b):Maximum wavelength is: ",round(lmax,1),"pm"
print"(c):The maximum KE of recoil electrons is:%.3g"%KEmax,"J"
print"which is equal to ",round(KEmax/1.6021773e-16,1),"keV(approx)"
#Variable declaration
c=3.0*10**8 #velocity of light, m/s
v= 0.5*c #velocity of electron and positron, m/s
y= 1.0/math.sqrt(1.0-(v/c**2)) #gamma, for relativistic momentum
m=0.511/c**2 #MeV/c**2
#Calculation
p1_p2=((2*m*c*c**2)*(v/c**2))/(math.sqrt(1-(v/c)**2))
p1_plus_p2=(2.0*m*(c**2))/math.sqrt(1.0-(v/c)**2.0)
p1=(p1_p2+p1_plus_p2)/2.0
p2=p1_plus_p2-p1
E1=p1 #MeV
E2=p2 #MeV
#Result
print"The Energy of first photon is,E1=",round(E1,3),"MeV"
print"The Energyof second photon is:,E2=",round(E2,3),"MeV"
print"NOTE:There is a mistake in the formula given in the book for p1_p2"
#Variable declaration
M= 4.9 #Linear attenuation coefficient for gamma ray in water, m**(-1)
I= 2.0 #Original intensity of gamma ray, MeV
#Calculation
#Part (a)
x= 10.0 #distance travelled under water, cm
x= x/100.0 #converting to m
Irel= math.exp(-(M*x)) #Relative intensity
#Part(b)
Ip= I/100 #Present intensity, 1 percent of Original, MeV
x2= math.log(I/Ip)/M #distance travelled, m
#Result
print"(a)Relative intensity of the beam is: ",round(Irel,2)
print"(b)The distance travelled by the beam is:",round(x2,2),"m"
#Variable declaration
H= 22.5 #Height of fall, m
F= 7.3*(10**14) #Original frequency, Hz
c= 3*(10**8) #velocity of light, m/s
g= 9.8 #Acceleration due to gravity, m/s**2
#Calculation
Frel= g*H*F/(c**2) #Change in frequency, Hz
#Result
print"The change in frquency of a photon fallin through 22.5 m is: ",round(Frel,1), "Hz"