Mass — Question 7

Question 7

Counting particles and adding their masses use different weights

The unified atomic mass unit, symbol u, is one twelfth of the mass of a free carbon-12 atom at rest in its ground state. For calculations here, use the approximate conversion $$1\ \mathrm{u}\approx1.66054\times10^{-27}\ \mathrm{kg}.$$ Consider an idealized collection of noninteracting particles of two types, with assigned masses $m_A=20\ \mathrm{u}$ and $m_B=22\ \mathrm{u}$. These are model particle masses, not tabulated masses of specific isotopes. Masses add, with binding effects excluded.

The total count is $N=2.00\times10^{20}$. Exactly 75% of the particles are A and 25% are B. Treat the count and fractions as exact model inputs; kilogram results inherit the approximate unit conversion.

1. Find the number of each type, the mean mass per particle in u, and the total mass in kilograms and milligrams. Use $1\ \mathrm{mg}=10^{-6}\ \mathrm{kg}$.

2. What fraction of the total mass is contributed by B? Explain why this is not the same as its 25% number fraction.

3. Does the mean mass require that some individual particle have that mass? Explain in this two-type model.

4. A second collection consists only of A particles and has exactly the same total mass as the first. Find its particle count relative to $N$ and its numerical count.

5. Is the carbon-12 definition of u equivalent to an exact kilogram conversion for u? Distinguish an exact mass ratio from the experimentally determined relation to the SI kilogram.

Question 7 - Solution

Counting particles and adding their masses use different weights

(a) Weight particle masses by particle counts.

$$N_A=0.75N=\boxed{1.50\times10^{20}},\qquad N_B=0.25N=\boxed{5.00\times10^{19}}.$$ The mean mass is $\overline m=0.75(20)+0.25(22)=\boxed{20.5\ \mathrm{u}}$. Hence $$M=N\overline m=(2.00\times10^{20})(20.5)\ \mathrm{u} =4.10\times10^{21}\ \mathrm{u},$$ $$M\approx(4.10\times10^{21})(1.66054\times10^{-27})\ \mathrm{kg} =\boxed{6.808214\times10^{-6}\ \mathrm{kg}=6.808214\ \mathrm{mg}}.$$ The last digits reflect the supplied approximate conversion, not an exact kilogram mass.

(b) Mass fraction includes the mass of each particle.

$$w_B=\frac{N_Bm_B}{N\overline m} =\frac{0.25(22)}{20.5} =\boxed{\frac{11}{41}\approx0.268293}.$$ B supplies about 26.8293% of the mass. Its particles are heavier than A particles, so its share of mass is greater than its share of the count.

(c) An average need not be an individual value.

$$\overline m=\frac{N_Am_A+N_Bm_B}{N_A+N_B}=20.5\ \mathrm u\notin\{20\ \mathrm u,22\ \mathrm u\}.$$

No particle in this model has mass 20.5 u: each is either 20 u or 22 u. The mean describes total mass divided by total count. It does not introduce a third particle type or split individual particles.

(d) The lighter pure collection needs more particles.

Equal total masses require $N'(20\ \mathrm{u})=N(20.5\ \mathrm{u})$, giving $$\boxed{N'/N=1.025},\qquad \boxed{N'=2.05\times10^{20}}.$$ The common unit cancels, so this count ratio does not depend on the approximate kilogram conversion.

(e) A defined ratio is not a fixed kilogram value.

The ratio of one u to the specified carbon-12 atom’s mass is exactly $1/12$. The atom’s mass expressed in kilograms is determined experimentally under the current SI. Thus u is not assigned an exact kilogram value by that ratio definition, and the approximate conversion supplied here must not be treated as exact.