# Chapter 15: Nuclear Models¶

## Example 15.2, Page 516¶

In [1]:
#Variable declaration
c=3*10**8#m/s
k=500.#Mev

#Calculations
p=(k)/(c*6.2*10**12)
h=6.63*10**-34#in j-s
lamda=h/p
r=lamda/angle

#Results
print "The wavelength is  %.1e  m"%lamda
print "The angle is  %.1e  m"%r
#answers vary due to rounding-off errors

The wavelength is  2.5e-15  m
The angle is  4.7e-15  m


## Example 15.4, Page 522¶

In [2]:
#Variable declaration
kb=4.44#in Mev
ka=7.70#in Mev
mb=1.
mB=17
ma=4.

#Calculations&Results
Q=(kb*(1+(mb/mB)))-(ka*(1-(ma/mB)))
#Part a
print "The value of Q is %.2f  Mev"%Q
c=3*10**8#m/s
m=Q/(931.5)
print "The atomic mass of Q is %.2e  u"%m

The value of Q is -1.19  Mev
The atomic mass of Q is -1.27e-03  u


## Example 15.5, Page 523¶

In [3]:
#Variable declaration
M_He=4.0026033#*u, Mass of helium
M1H1=1.00782525#*u, electron mass
Mon1=1.0086654#*u, neutron mass

#Calculations
Mass=(2*M1H1)+(2*Mon1)
delta_M=(Mass)-M_He

#Result
print "The binding energy of helium is  %f*u"%delta_M

The binding energy of helium is  0.030378*u


## Example 15.8, Page 531¶

In [4]:
#Variable declaration
N=0.60

#Calculations&Results
rho=(N)/((4/3))
print "The density is  %.2f  /pi*a**3"%rho
h=6.63*10**-34#in j-s
a=1.1#F
M=1
ef=43#in Mev
En=7#in Mev
Vo=ef+En
print "The depth of the net nuclear potential acting on neutron is %.f Mev"%Vo

The density is  0.60  /pi*a**3
The depth of the net nuclear potential acting on neutron is 50 Mev