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Find the binding energy of a nucleus cinsisting of equal numbers of protons and neutrons and having the radius one and a half times smaller than that of Al^(27) nucleus. |
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Answer» Solution :The mass number of the given nucleus MUST be `27//((3)/(2))^(3)=8` THUS the nucleus is `Be^(8)`. Then the BINDING energy is `E_(b)= 4xx0.00867+4+x0.00783-0.00531 am u` `=0.06069am u=56.5MeV` On using `1 am u= 931 MeV` |
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