Solve qus no. 5:

5. An alpha particle having kinetic energy of 10 MeV is moving towards a stationary nucleus of atomic number 50. The distance of closest approach will be.
(a) 1.44 x 10-15 m          (b) 1.44 x 10-15 m
(c) 1.44 x 10-13 m           (d)  14.4 x 10-16 m

 

Dear student,
KE=10MeV charge on nucleus=50e  Clet the alpha particle be at infinity charge on alpha particle =2eKE will be convert to PE at maximum approach 107eV=K100e2rV1=9×104×1.6×10-19rr=1.44×10-14 m Regards
 

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