1)the energy of an alpha particle is 1.2×10^-13J.up to what closet distance can it reach the nucleus of silver(Z=47)?
Ans is 1.8×10^-13.

Energy of the alpha particle, E=1.2×10-13 JAtomic number of silver, Z=47Distance of closest approach is given byr=Ze2e4π0×12mv2r=9×109×47×21.6×10-1921.2×10-13r=1804.8×10-16 mr=1.8×10-13 m

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