Medium
Estimate the average mass density of a sodium atom assuming its size to be about $2.5 \; \mathring{A}$. (Use the known values of Avogadro's number and the atomic mass of sodium). Compare it with the mass density of sodium in its crystalline phase: $970 \; kg \; m^{-3}$. Are the two densities of the same order of magnitude? If so,why?
(A) Diameter of sodium atom $= 2.5 \; \mathring{A} = 2.5 \times 10^{-10} \; m$.
Radius of sodium atom,$r = 1.25 \times 10^{-10} \; m$.
Volume of one sodium atom,$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.14 \times (1.25 \times 10^{-10})^3 \approx 8.18 \times 10^{-30} \; m^3$.
Mass of one sodium atom,$m = \frac{\text{Atomic mass}}{\text{Avogadro's number}} = \frac{23 \times 10^{-3} \; kg}{6.023 \times 10^{23}} \approx 3.82 \times 10^{-26} \; kg$.
Density of sodium atom,$\rho = \frac{m}{V} = \frac{3.82 \times 10^{-26}}{8.18 \times 10^{-30}} \approx 4.67 \times 10^3 \; kg \; m^{-3}$.
Given density of crystalline sodium $= 970 \; kg \; m^{-3} \approx 10^3 \; kg \; m^{-3}$.
Both densities are of the order of $10^3 \; kg \; m^{-3}$. They are of the same order of magnitude because in the solid phase,atoms are closely packed,but the inter-atomic spacing is comparable to the atomic size.
Radius of sodium atom,$r = 1.25 \times 10^{-10} \; m$.
Volume of one sodium atom,$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.14 \times (1.25 \times 10^{-10})^3 \approx 8.18 \times 10^{-30} \; m^3$.
Mass of one sodium atom,$m = \frac{\text{Atomic mass}}{\text{Avogadro's number}} = \frac{23 \times 10^{-3} \; kg}{6.023 \times 10^{23}} \approx 3.82 \times 10^{-26} \; kg$.
Density of sodium atom,$\rho = \frac{m}{V} = \frac{3.82 \times 10^{-26}}{8.18 \times 10^{-30}} \approx 4.67 \times 10^3 \; kg \; m^{-3}$.
Given density of crystalline sodium $= 970 \; kg \; m^{-3} \approx 10^3 \; kg \; m^{-3}$.
Both densities are of the order of $10^3 \; kg \; m^{-3}$. They are of the same order of magnitude because in the solid phase,atoms are closely packed,but the inter-atomic spacing is comparable to the atomic size.
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View SolutionMatch the $\text{LIST-I}$ with $\text{LIST-II}$:
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| $\text{LIST-I}$ | $\text{LIST-II}$ |
|---|---|
| $A$. Boltzmann constant | $I$. $ML^2 T^{-1}$ |
| $B$. Coefficient of viscosity | $II$. $MLT^{-3} K^{-1}$ |
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| $D$. Thermal conductivity | $IV$. $ML^{-1} T^{-1}$ |
Choose the correct answer from the options given below:
Match List-$I$ with List-$II$.
List-$I$ List-$II$ $(a)$ Capacitance,$C$ $(i)$ $M^{1} L^{1} T^{-3} A^{-1}$ $(b)$ Permittivity of free space,$\varepsilon_{0}$ $(ii)$ $M^{-1} L^{-3} T^{4} A^{2}$ $(c)$ Permeability of free space,$\mu_{0}$ $(iii)$ $M^{-1} L^{-2} T^{4} A^{2}$ $(d)$ Electric field,$E$ $(iv)$ $M^{1} L^{1} T^{-2} A^{-2}$
Choose the correct answer from the options given below:
| List-$I$ | List-$II$ |
| $(a)$ Capacitance,$C$ | $(i)$ $M^{1} L^{1} T^{-3} A^{-1}$ |
| $(b)$ Permittivity of free space,$\varepsilon_{0}$ | $(ii)$ $M^{-1} L^{-3} T^{4} A^{2}$ |
| $(c)$ Permeability of free space,$\mu_{0}$ | $(iii)$ $M^{-1} L^{-2} T^{4} A^{2}$ |
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Choose the correct answer from the options given below:
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$I$ (Fundamental forces in nature) $II$ (Relative strength) $(A)$ Strong nuclear force $(e)$ $10^{-2}$ $(B)$ Weak nuclear force $(f)$ $1$ $(C)$ Electromagnetic force $(g)$ $10^{10}$ $(D)$ Gravitational force $(h)$ $10^{-13}$ $(i)$ $10^{-39}$
The correct match is:
| $I$ (Fundamental forces in nature) | $II$ (Relative strength) |
|---|---|
| $(A)$ Strong nuclear force | $(e)$ $10^{-2}$ |
| $(B)$ Weak nuclear force | $(f)$ $1$ |
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Medium
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