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Wave Optics

Question
CBSEENPH12038709

Calculate the force between two fission fragments of equal masses and sizes that are produced in the fission of Pu94239 (by a thermal neutron) in which 4 neutrons are emitted. Given R0 = 1.2 fm. 

Solution

The undertaking fission process can be wriiten as : 

              Pu94239 + n01  2 X47118 + 4 n01 + Q 

Total mass number of Pu94239 and neutron = 240 

Mass number of each fragment = 240-42 = 118

Atomic number of each fragment = 942 = 47 

Radius of each nucleus formed by the fission of Pu94239 is, 

                       R = R0 A1/3 

                          = 1.2 × 10-15 × (118)1/3 = 5.886 × 10-15m 

Distance between the centre of the two fragments = 2 × 5.886 ×10-15 m = 11.77 × 10-15m 

Electrostatic force between them is given by Coulomb's law, 
                        F = 14πεoq1q2r2 

Therefore, 

      F = 14πε0(47×1.6×10-19)2(11.77 ×10-15)2 = 3.763 × 103N.

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