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Assuming the radius of a nucleus to be equal to R= 0.13 root(3)(A) pm, where A is its mass number, evalulate the density of nuclei and the number of nucleons per unit volume of the nucleus. |
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Answer» Solution :This problem has misprint. Actually the radius `R` of a nucleus is given by `R=1.3root(3)(A)FM` where `fm= 10^(-15)m` Then the number of nucleous per unit VOLUME is `(A)/((4 pi)/(3)R^(3))=(3)/(4PI)XX(1.3)^(-3)xx10^(+39)CM^(-3)= 1.09xx10^(38)`per cc The corresponding mass density is `(1.09xx10^(38)xx"mass of nucleon") per c c =1.82xx10^(11)kg//c c` |
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