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The order of magnitude of density of urn...

The order of magnitude of density of urnaitum nucleus is:
`(m_("nucleus") = 1.67xx10^(-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)`

Text Solution

Verified by Experts

The correct Answer is:
B
Nuclear density `= ("Nuclear mass")/("Nuclear volume")`
`rArr" "rho = (mA)/((4)/(3) pi r_(0)^(3)A) = (3 m)/(4 pi r_(0)^(3))`
where A is the mass number.
It is independent of A. So we can use the mass of proton instead of mass of uranium and `r_(0) = 1.2 xx 10^(-15)`m.
`therefore" "rho = (3 xx 1.67 xx 10^(-27))/(4 xx 3.142 xx (1.2 xx 10^(-15))^(3)) = 2.3 xx 10^(17) kg//m^(3)`
`rho` is of the order of `10^(17) kg//m^(3)`
[It is very large, as compared to the density of ordinary matter like water. `rho = 10^(3) k//m^(3)`]
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