# 10: Quantum Physics and Schrodinger wave equation¶

## Example number 10.1, Page number 258¶

In :
#importing modules
import math
from __future__ import division

#Variable declaration
me=9.1*10**-31;   #mass of electron(kg)
h=6.625*10**-34;   #planck's constant(Jsec)
deltax=10**-8;    #uncertainity in position(m)

#Calculation
deltap=(h/(2*math.pi*deltax));   #uncertainity principle(kgm/sec)
deltav=(deltap/me);   #minimum uncertainity in velocity(m/sec)

#Result
print "minimum uncertainity in velocity is",round(deltav/10**5,3),"*10**5 m/sec"

minimum uncertainity in velocity is 0.116 *10**5 m/sec


## Example number 10.2, Page number 259¶

In :
#importing modules
import math
from __future__ import division

#Variable declaration
lamda=0.2865*10**-10;   #wavelength(m)
mp=1.67*10**-27;   #mass of proton(kg)
h=6.625*10**-34;   #planck's constant(Jsec)
q=1.6*10**-19;     #charge of proton(C)

#Calculation
v=(h/(mp*lamda));    #velocity(m/sec)
KE=0.5*mp*(v**2);    #kinetic energy of proton(J)
KE=KE/q;     #kinetic energy of proton(eV)

#Result
print "kinetic energy of proton is",int(KE),"eV"

kinetic energy of proton is 1 eV


## Example number 10.3, Page number 259¶

In :
#importing modules
import math
from __future__ import division

#Variable declaration
KE=0.025;   #kinetic energy of neutron(eV)
q=1.6*10**-19;     #charge of proton(C)
mn=1.676*10**-27;  #mass of neutron(kg)
h=6.625*10**-34;   #planck's constant(Jsec)
me=9.1*10**-31;    #mass of electron(kg)
c=3*10**8;    #velocity of light(m/s)

#Calculation
KE=KE*q;   #kinetic energy of neutron(J)
v=math.sqrt((2*KE)/mn);   #velocity(m/s)
lamdan=h/(mn*v);   #debroglie wavelength of neutron(m)
p=(h/lamdan);   #momentum of electron and photon(kgm/s)
ve=(p/me);   #velocity of electron(m/s)
Ee=0.5*p*ve;  #energy of electron(J)
Ee=Ee/q;    #energy of electron(eV)
Ep=h*c/lamdan;  #energy of photon(J)
Ep=Ep/q;       #energy of photon(eV)

#Result
print "wavelength of beam of neutron is",round(lamdan*10**10,3),"angstrom"
print "momentum of electron and photon is",p,"kgm/s"
print "energy of electron is",round(Ee,2),"eV"
print "energy of photon is",round(Ep/10**3,2),"*10**3 eV"
print "answers in the book vary due to rounding off errors"

wavelength of beam of neutron is 1.809 angstrom
momentum of electron and photon is 3.66169359723e-24 kgm/s
energy of electron is 46.04 eV
energy of photon is 6.87 *10**3 eV
answers in the book vary due to rounding off errors


## Example number 10.4, Page number 260¶

In :
#importing modules
import math
from __future__ import division

#Variable declaration
e=1.6*10**-19;    #charge of electron(C)
V=200;    #potential difference(V)
lamda=0.0202*10**-10;   #debroglie wavelength(m)
h=6.625*10**-34;    #planck's constant(Jsec)

#Calculation
#eV=0.5*m*(v^2)
#mv=sqrt(2*m*eV)
m=((h**2)/(2*(lamda**2)*e*V));    #mass of particle(kg)

#Result
print "mass of particle is",m,"kg"
print "hence it is a proton"

mass of particle is 1.68069555834e-27 kg
hence it is a proton


## Example number 10.5, Page number 260¶

In :
#importing modules
import math
from __future__ import division

#Variable declaration
mn=1.676*10**-27;   #mass of neutron(kg)
e=1.6*10**-19;    #charge of electron(C)
h=6.622*10**-34;   #planck's constant(Jsec)

#Calculation
E=e;     #energy of neutron(J)
v=math.sqrt((2*E)/mn);   #velocity of neutron(m/sec)
lamda=(h/(mn*v));   #de-broglie wavelength(m)

#Result
print "de-broglie wavelength of neutron is",round(lamda*10**10,3),"angstrom"

de-broglie wavelength of neutron is 0.286 angstrom


## Example number 10.6, Page number 261¶

In :
#importing modules
import math
from __future__ import division

#Variable declaration
h=6.625*10**-34;    #planck's constant(Jsec)
c=3*10**8;    #velocity of light(m/s)
mo=9.1*10**-31;   #rest mass of particle(kg)
q=1.6*10**-19;   #charge of electron(C)

#Calculation
#acc. to uncertainity principle delx*delp >= (h/2*%pi)
deltax=2*r;   #uncertainity in position(m)
deltap=(h/(2*math.pi*deltax));    ##uncertainity in momentum
#from einstein's relavistic relation E=mc2=KE+rest mass energy=0.5mv2+moc2
#when velocity of particle is very high
#m=(mo/sqrt(1-((v/c)^2))) where m-mass of particle with velocity v,mo-rest mass of particle, c-velocity of particle
p=deltap    #assume
E=math.sqrt(((p*c)**2)+((mo*(c**2))**2));    #energy(J)
E=E/q;      #energy(eV)

#Result
print "energy is",round(E/10**6),"MeV"
print "this value is much higher than experimentally obtained values of energy of electron of a radioactive nuclei i.e 4 Mev this proves that electron cannot reside within nucleus"

energy is 10.0 MeV
this value is much higher than experimentally obtained values of energy of electron of a radioactive nuclei i.e 4 Mev this proves that electron cannot reside within nucleus