In [3]:

```
#initiation of variable
from math import sqrt
h=6.6*10**-34; # h(planck's constant)= 6.6*10^-34
m1= 10.0**3;v1=100.0; # for automobile
#calculation
w1= h/(m1*v1); # ['w'-wavelength in metre'm'-mass in Kg 'v'-velocity in metres/sec.] of the particles
m2=10.0*(10**-3);v2= 500; # for bullet
w2=h/(m2*v2);
m3=(10.0**-9)*(10.0**-3); v3=1.0*10**-2;
w3=h/(m3*v3);
m4=9.1*10**-31;k=1*1.6*10**-19; # k- kinetic energy of the electron & using 1ev = 1.6*10^-19 joule
p=sqrt(2.0*m4*k); # p=momentum of electron ;from K=1/2*m*v^2
w4=h/p;
hc=1240;pc=100 # In the extreme relativistc realm, K=E=pc; Given pc=100MeV,hc=1240MeV
w5= hc/pc;
#result
print "Wavelength of the automobile in m is",w1;
print "Wavelength of the bullet in m is ",w2 ;
print"Wavelength of the smoke particle in m is",w3 ;
print "Wavelength of the electron(1ev) in nm is",round(w4*10**9,3) ;
print "Wavelength of the electron (100Mev) in fm is",w5;
```

In [4]:

```
#initiation of variable
from math import pi
# w=wavelength; consider k=2*(pi/w);
# differentiate k w.r.t w and replace del(k)/del(w) = 1 for equation.4.3
# which gives del(w)= w^2 /(2*pi*del(x)), hence
w=20; delx=200; # delx=200cm and w=20cm
#calculation
delw=(w**2)/(delx*2*pi);
#result
print "Hence uncertainity in length in cm is",round(delw,3);
```

In [6]:

```
#initiation of variable
from math import pi
delt=1.0; #consider time interval of 1 sec
delw=1/delt; # since delw*delt =1 from equation 4.4
delf=0.01 #calculated accuracy is 0.01Hz
#calculation
delwc =2*pi*delf # delwc-claimed accuracy from w=2*pi*f
#result
print "The minimum uncertainity calculated is 1rad/sec. The claimed accuracy in rad/sec is \n",round(delwc,3);
print "thus there is a reason to doubt the claim"
```

In [4]:

```
#initiation of variable
from math import pi
m=9.11*10**-31;v=3.6*10**6; #'m','v' - mass an velocity of the electron in SI units
h=1.05*10**-34; #planck's constant in SI
p=m*v; #momentum
delp=p*0.01; #due to 1% precision in p
delx = h/delp; #uncertainity in position
#result
print "Uncertainity in position in nm is",round(delx*10**9,2);
#partb
print "Since the motion is strictly along X-direction, its velocity in Y direction is absolutely zero.\n So uncertainity in velocity along y is zero=> uncertainity in position along y is infinite. \nSo nothing can be said about its position/motion along "
```

In [5]:

```
#initiation of variable
from math import pi
m=0.145;v=42.5; #'m','v' - mass an velocity of the electron in SI units
h=1.05*10**-34; #planck's constant in SI
p=m*v; #momentum
delp=p*0.01;#due to 1% precision in p
#calculation
delx = h/delp#uncertainity in position
#result
print "Uncertainity in position is %.1e" %delx;
#part b
print "Motion along y is unpredictable as long as the veloity along y is exactly known(as zero).";
```

In [14]:

```
#initiation of variable
from math import sqrt
mc2=2.15*10**-4; #mc2 is the mass of the electron, concidered in Mev for the simplicity in calculations
hc=197.0 # The value of h*c in Mev.fm for simplicity
delx= 10.0 # Given uncertainity in position=diameter of nucleus= 10 fm
#calculation
delp= hc/delx ; #Uncertainiy in momentum per unit 'c' i.e (Mev/c) delp= h/delx =(h*c)/(c*delx);hc=197 Mev.fm 1Mev=1.6*10^-13 Joules')
p=delp; # Equating delp to p as a consequence of equation 4.10
K1=p**2+mc2**2 # The following 3 steps are the steps invlolved in calculating K.E= sqrt((p*c)^2 + (mc^2)^2)- m*c^2
K1=sqrt(K1)
K1= K1-(mc2);
#result
print "Kinetic energy was found out to be in Mev is", round(K1,3)
```

In [9]:

```
#initiation of variable
h=6.58*10**-16; # plack's constant
delt1=26.0*10**-9;E1=140.0*10**6 #given values of lifetime and rest energy of charged pi meson
delt2=8.3*10**-17;E2=135.0*10**6; #given values of lifetime and rest energy of uncharged pi meson
delt3=4.4*10**-24;E3=765*10**6; #given values of lifetime and rest energy of rho meson
#calculation
delE1=h/delt1; k1=delE1/E1; # k is the measure of uncertainity
delE2=h/delt2; k2=delE2/E2;
delE3=h/delt3; k3=delE3/E3;
#result
print "Uncertainity in energy of charged pi meson is %.1e" %k1;
print "Uncertainity in energy of uncharged pi meson is %.1e" %k2;
print "Uncertainity in energy of rho meson is ",round(k3,2);
```

In [17]:

```
#initiation of variable
h=1.05*10**-34; #value of planck's constant in J.sec
delx= 1.0; # uncertainity in positon= dimension of the ball
delp=h/delx; # uncertainity in momentum
m=0.1; #mass of the ball in kg
#calculation
delv=delp/m; # uncertainity in velocity
#result
print "The value of minimum velocity was found out to be in m/sec",delv;
```