Saved Bookmarks
| 1. |
The nuclear mass of ""_(26)^(56)"Fe"is 55.85 amu. Calculate itsnuclear density. |
|
Answer» Solution :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)` |
|