Given the mass of iron nucleus …
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Gaurav Seth 6 years, 9 months ago
According to rutherford's relationsradius of nucleus = ro×A13 ro = constant = 1.2×10−15 m
therefore
r = 1.2×10−15 ×56−−√3
r = 4.59 ×10−15 m
also,
1u = 1.67×10−27 kg
55.85 u = 9.327 ×10−26 kg = mass of nucleus of iron atom
nuclear density = mass of nucleus of iron atomvolume of nucleus
nuclear density = 9.327 ×10−26 kg43π×r3 =9.327 ×10−26 kg43π×(4.59 ×10−15 )3
nuclear density = 4.851×1018 kg/m^3
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