NEET Class neet Answered
Since the positive charge is confined to the nucleus,the actual size of the nucleus has to be less than 4.0×10-14m.please explain .how it is related to 4.0×10-14m
Asked by valavanvino1011 | 10 Apr, 2019, 10:55: AM
Expert Answer
Experiments of scattering of α-particle from thin gold foil was performed by Geiger and Marsden as per Rutherford's suggestion.
This experiment revealed that the distance of closest approach to a gold nucleus of an α-particle of kinetic energy 5.5 MeV is about 4.0×10-14 m.
This value of closest approach distance was arrived by assuming that the coulomb repulsive force was fully responsible for scattering.
Since the positive charge is confined to the nucleus, the actual size of the nucleus has to be less than 4.0 × 10–14 m.
Answered by Thiyagarajan K | 10 Apr, 2019, 01:27: PM
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