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Order of magnitude of density of uranium nucleus is , [m = 1.67 xx 10^(-27 kg]`A. `10^(20) kg//m^(3)`B. `10^(17) kg//m^(3)`C. `10^(14) kg//m^(3)`D. `10^(11) kg//m^(3)` |
Answer» Correct Answer - B Radius of a nucleus is given by `R=R_(0)A^(1//3)` (where `R_(0)=1.25xx10^(-15) m)` `=1.25 A^(1//3)xx10^(-15) m` Here, A is the mass number and mass of the uranium nucleus will be `m~~Am_(p)` `m_(p)=` mass of proton `=A(1.67xx10^(-27) kg)` `:.` Density `rho=("mass")/("Volume")=m/(4/3 pi R^(3))=(A(1.67xx10^(-27)kg))/(A(1.25xx10^(-15) m)^(3))` `implies rho~~2.0xx10^(17) kg//m^(3)` |
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