What is the density of a bar of silver with a mass of 68 g and a volume of 6.5 cm3?
1 answer:
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Answer:
See Explanation
Explanation:
Hence the mass defect is;
[235.04393 + 1.00867] - [ 136.92532 + 96.91095 + 2(1.00867)]
= 236.0526 - 235.85361
= 0.19899 amu
Since 1 amu = 1.66 * 10^-27 Kg
0.19899 amu = 0.19899 * 1.66 * 10^-27 = 3.3 * 10^-28 Kg
Binding energy = Δmc^2
Binding energy = 3.3 * 10^-28 Kg * (3 * 10^8)^2 = 2.97 * 10^-11 J
ii)
Hence the mass defect is;
[10.01294 + 1.00867] - [7.01600 + 4.00260]
= 11.02161 - 11.0186
= 0.00301 amu
Since 1 amu = 1.66 * 10^-27 Kg
0.00301 amu = 0.00301 * 1.66 * 10^-27 = 4.997 * 10^-30 Kg
Binding energy = Δmc^2
Binding energy = 4.997 * 10^-30 Kg * (3 * 10^8)^2 = 4.5 * 10^-13 J