CATAGORY: Electricity & Magnetism
QUESTION: A proton is approaching an iron nucleus with an initial speed of
5.05 x 106  m/s, when the proton is 20.0 nuclear radii from the iron
(center to center distance).

a) how far from the nucleus will the proton be when its speed is 2.03 x
10 m/s?  Answer in units of the iron's radius (which is approximately 5.8
x 10-15 m).
b) How close to the nucleus will the proton get?  Answer in units of the
nucleus' radius.

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Use conservation of total energy to solve this problem.

(1) KE(1) + PE(1) = KE(2) + PE(2)

where (1) means 'initial' and (2) means 'final'.
Now 

(2) PE = keZe/r = kZe^2/r

where k = 8.9875 x 10^9 Nm^2/C^2, e = 1.602 x 10^-19 C and
Z(for iorn) is 26, and r is the radius.

(3) KE = (1/2)mv^2

where m is the proton mass 1.67 x 10^-27kg , and v is the velocity.

thus we have

(4) (1/2)mv(1)^2 + kZe^2/r1 = (1/2)mv(2)^2 + kZe^2/r2

now r1 = 20ro = 20(5.8 x 10-15 m) 

solving for r2 we have

(5) (1/2)m(v(1)^2 - v(2)^2) + kZe^2/r1 = kZe^2/r2

or
                       kZe^2
(6)   r2 = -----------------------------------
           (1/2)m(v(1)^2 - v(2)^2) + kZe^2/r1
or           
           
                      r1 kZe^2
(6)   r2 = -----------------------------------
           (1/2)mr1(v(1)^2 - v(2)^2) + kZe^2
or
                      r1 
(7)   r2 = ----------------------------
           (1/2)m(v(1)^2 - v(2)^2) 
           ----------------------- + 1         
                     kZe^2
                     -----
                      r1
          
or

                       
                   20(5.8 x 10-15 m)            
   = ----------------------------------------------------------
        (1/2)1.67 x10^-27kg((5.05x10^6 m/s)^2-(2.03x10^6m/s)^2)
    1 + -------------------------------------------------------
         8.9875 x 10^9 Nm^2/C^2(26)(1.602 x 10^-19 C)^2
         ---------------------------------------------
                     20(5.8 x 10-15 m)
                  
                   
                       
       20(5.8 x 10-15 m)             
   = -------------------
        1.785 x 10^-14J
    1 + ----------------
        5.17 x 10^-14J 
        
                       
       20(5.8 x 10-15 m)             
   = -------------------
        1.79 x 10^-14J
    1 + ----------------
        5.17 x 10^-14J
        
        
   =  14.9(5.8x10^-15m)
   
   
   =  8.62 x 10^-14 m
   
the closest approach is when v2 = 0. Substitution into (7) gives
   
                 r1 
(7)   r2 = ----------------
           (1/2)mv(1)^2 
           ------------ + 1         
               kZe^2
               -----
                 r1
          
or

                       
                   20(5.8 x 10-15 m)            
   = -------------------------------------------------
           (1/2)1.67 x10^-27kg (5.05x10^6 m/s)^2
    1 + ----------------------------------------------
         8.9875 x 10^9 Nm^2/C^2(26)(1.602 x 10^-19 C)^2
         ---------------------------------------------
                     20(5.8 x 10-15 m)
                  
                   
                       
       20(5.8 x 10-15 m)             
   = -------------------
        2.13 x 10^-14J
    1 + ----------------
        5.17 x 10^-14J 
        
   
   
   = 14.16(5.8 x 10^-15m)
   
   
   *************************************
   *                                   *
   * (a) r2 = 14.9ro = 8.62 x 10^-14m  *
   *                                   *
   * (b) r2 = 14.16ro = 8.21 x 10^-14m *
   *                                   *
   *************************************