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The nuclear mass of ""(26)^(56)"Fe" is ...

The nuclear mass of `""_(26)^(56)"Fe"` is `55.85` amu. Calculate its nuclear density.

Text Solution

Verified by Experts

Here `M_(Fe) = 55.85 amu = 55.85 xx 1.66 xx 10^(-27)` kg
` = 9.27 xx 10^(-26)` kg
Nuclear radius = `R_(0)A^(1/3) = 1.1 xx 10^(-15) xx(56)^(1/3) m`
`rho_(nu)` = (Nuclear Mass)/(Nuclear Volume) =`(M_(Fe))/4/3 pi R^(3)`
` = (9.27 xx 10^(-26))/(4/3 xx 3.14 xx1.1 xx 10^(-15) xx 56)`
` (9.27 xx 10^(-26))/(4/3 xx 3.14 xx(1.1 xx 10^(-15))^(3) xx 56)`
` = (9.27 xx 10^(-26))/(283.688 xx 10^(-45) xx 1.1) = (9.27 xx 10^(+19))/(312.057)`
`rho_(nu) = 2.9 xx 10^(17) kg m^(-3)`
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