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InterviewSolution
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Calucate the density of the nucelus of `._(47)^(107)Ag` assuming `R_("nucleus")` is `1.4A^(1//3)xx 10^(-13) cm`. Where `A` is mass number of nucelsus. Compare its density with density of metallic silver `10.5g cm^(-3)` |
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Answer» Volume of `._(47)Ag^(107)` nucleus `= (4)/(3)pi r^(3)` `= (4)/(3) xx (2)/(7) xx [1.4xx(107)^(1//3) xx 10^(-13)]^(3)` `= 1.23xx10^(-36) cm^(3)` `:.` Density of nucleus `= (m)/(V)` `= (107)/(6.023xx10^(23)xx1.23xx10^(-36))` `= 1.445xx10^(14)g//cm^(3)` Thus `("Density of nucleus")/("Density of atom") = (1.445xx10^(14))/(10.5)` `= 1.38xx10^(13)` |
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