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 Share with your friends Share 4 Decoder_2 answered this 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 8 View Full Answer Mahi answered this Hope this will help you -8