Kinetic Theory Class 11 NCERT Solutions: Download Topic-wise Questions & Answers PDF

• Introduction to Kinetic Theory
• Molecular nature of matter
• Behaviour of gases
• Kinetic theory of an ideal gas
• Law of equipartition of energy
• Specific heat capacity
• Mean free path

NCERT Solution for Class XI Physics Chapter Kinetic Theory is available here in PDF format also. The candidates can click on the Physics NCERT Class 11 PDF link for the Kinetic Theory chapter given below and download it. The PDF format of solutions can be downloaded and used by the students to get their doubts cleared instantly.

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Q.13.1 Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.

Ans.13.1

Diameter of the oxygen molecule, d = 3Å

Radius, r =  $\frac{d}{2}$ = 1.5 Å = 1.5  $\times {10}^{-8}$ cm

Actual volume occupied by 1 mole of oxygen gas at STP = 22.4 lit = 22400  ${cm}^3$

Molecular volume of oxygen gas, V =  $\frac{4}{3}\pi r^3$ N,

where N = Avogadro’s number = 6.023  $\times {10}^{23}$ molecules/mole

Therefore, V =  $\frac{4}{3}$  $\times 3.1416\times ({1.5\mathrm{}\times {10}^{-8})}^3$  $\times$ 6.023  $\times {10}^{23}$ = 8.51  ${cm}^3$

Ratio of the molecular volume to the actual volume of oxygen =  $\frac{8.51}{22400}$ = 3.8  $\times {10}^{-4}$

Q.13.2 Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres.

Ans.13.2

The ideal gas equation relating to pressure (P), volume (V) and absolute temperature (T) is given by the relation

PV = nRT, where

R is the universal gas constant = 8.314 J /mol/K

n = number of moles = 1 (given)       

T = standard temperature = 273 K

P = standard pressure = 1 atm = 1.013  $\times {10}^5$ N/  $m^2$

Therefore V =  $\frac{nRT}{P}$ =  $\frac{1\times 8.314\times 273}{1.013\mathrm{}\times {10}^5}$ = 0.02240  $m^3$ = 22.4 litres

Hence the molar volume of a gas at STP is 22.4 litres.

Q.13.3 Figure 13.8 shows plot of PV/T versus P for 1.00×10–3 kg of oxygen gas at two different temperatures.

(a) What does the dotted plot signify?

(b) Which is true: T₁ > T₂ or T₁ < T₂?

(c) What is the value of PV/T where the curves meet on the y-axis?

(d) If we obtained similar plots for 1.00×10^–3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen

yields the same value of PV/T (for low pressure high temperature region of the plot) ? (Molecular mass of H₂ = 2.02 u, of O₂ = 32.0 u, R = 8.31 J mo1^–1 K^–1.)

Ans.13.3

The dotted plot in the graph signifies the ideal behaviour of the

gas, i.e.  $\frac{PV}{T}$ =  $\mu$ R, (where  $\mu$ is the number of moles and R is the universal gas constant) is a constant quantity. It is not dependent on the pressure of the gas. The curve of the gas at temperature  $T_1$ is closer to the dotted plot than the curve of the gas at temperature  $T_{2.}$

A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore  $T_1$ >  $T_2$ is true for the given plot.

The value of the ratio  $\frac{PV}{T}$ , where the two curves meet, is  $\mu$ R. This is because the ideal gas equation is given as  $\frac{PV}{T}$ =  $\mu$ R. From the given data, we have

Molecular mass of oxygen = 32 g

Mass of oxygen = 1  $\times {10}^{-3}$ = 1 g

R = 8.314 J/mole/K

Then  $\frac{PV}{T}$ =  $\frac{1}{32}\times 8.314$ = 0.26 J/k

Hence, the value of the ratio  $\frac{PV}{T}$ , where the curves meet on the y-axis, is 0.26 J/K.

If we obtain similar plots for 1.00  $\times {10}^{-3}$ kg of hydrogen, then we will not get the same values of  $\frac{PV}{T}$ at the point where the curves meet on the y-axis. This is because the molecular mass of hydrogen (2.02 u) is different from the oxygen (32.0 u). We have:

$\frac{PV}{T}=0.26\mathrm{}\mathrm{J}/\mathrm{K}$ .

Given, molecular mass (M) of  $H_2$ = 2.02 u

$\frac{PV}{T}$ =  $\mu$ R at constant temperature, where  $\mu$ =  $\frac{m}{M}$

m = mass of  $H_2$

Then, m =  $\frac{PV}{T}\times \frac{M}{R}$ =  $\frac{0.26\times 2.02}{8.31}$ = 6.3  $\times {10}^{-2}$ g = 6.3  $\times {10}^{-5}$ kg

Hence, 6.3  $\times {10}^{-5}$ kg of  $H_2$ will yield the same value of  $\frac{PV}{T}$

Q.13.4 An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol^–1 K^–1, molecular mass of O₂ = 32 u).

Ans.13.4

Volume of oxygen,  $V_1$ = 30 litres = 30  $\times {10}^{-3}{m}^3$

Gauge pressure,  $P_1$ = 15 atm = 15  $\times 1.013\times {10}^5$ Pa

Temperature,  $T_1$ = 27  $℃$ = 300 K

Universal gas constant, R = 8.314 J/mole/K

Let the initial number of moles of oxygen gas cylinder be  $n_1$

From the gas equation, we get  $P_1{V}_1=n_1\mathrm{R}{T}_1$

$n_1=\frac{P_1{V}_1}{\mathrm{R}{T}_1}$ =  $\frac{15\mathrm{}\times 1.013\times {10}^5\times 30\times {10}^{-3}}{8.314\times 300}$ = 18.276

But  $n_1$ =  $\frac{m_1}{M}$ , where  $m_1$ = initial mass of oxygen

M = Molecular mass of oxygen = 32 g

$m_1$ =  $n_1$ M = 18.276  $\times 32=584.84$ g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces, but volume remained unchanged.

Hence, Volume,  $V_2$ = 30 litres = 30  $\times {10}^{-3}{m}^3$

Gauge pressure,  $P_2$ = 11 atm = 11  $\times 1.013\times {10}^5$ Pa

Temperature,  $T_2$ = 17  $℃$ = 290 K

Let the final number of moles of oxygen left in the cylinder be  $n_2$

From the gas equation, we get  $P_2{V}_2=n_2\mathrm{R}{T}_2$

$n_2=\frac{P_2{V}_2}{\mathrm{R}2}$ =  $\frac{11\mathrm{}\times 1.013\times {10}^5\times 30\times {10}^{-3}}{8.314\times 290}$ = 13.864

But  $n_2$ =  $\frac{m_2}{M}$ , where  $m_2$ = remaining mass of oxygen

M = Molecular mass of oxygen = 32 g

$m_2$ =  $n_2$ M = 13.864  $\times 32=443.65$ g

So the mass of oxygen taken out =  $m_1-m_2$ = 141.19 gm = 0.141 kg

Q.13.5 An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?

Ans.13.5

Volume of the air bubble  $,$  $V_1$ = 1.0  ${cm}^3$ = 1  $\times {10}^{-6}$  $m^3$

Height achieved by bubble, d = 40 m

Temperature at a depth of 40 m,  $T_1$ = 12  $℃$ = 285 K

Temperature at the surface of the lake,  $T_2$ = 35  $℃$ = 308 K

The pressure at the surface of the lake,  $P_2$ = 1 atm = 1.013  $\times {10}^5$ Pa

The pressure at 40m depth:  $P_1$ = 1 atm + d  $\rho g$

Where  $\rho$ is the density of water =  ${10}^3$ kg/  $m^3$

G = acceleration due to gravity = 9.8 m/  $s^2$

Hence  $P_1$ = 1.013  $\times {10}^5+(40\times {10}^3\times 9.8)$ = 493300 Pa

From the relation  $\frac{P_1{V}_1}{T_1}$ =  $\frac{P_2{V}_2}{T_2}$ , where  $V_2$ is the volume of bubble when it reaches the surface, we get,  $V_2$ =  $\frac{P_1{V}_1{T}_2}{T_1{P}_2}$ =  $\frac{493300\times 1\times {10}^{-6}\times 308}{285\times 1.013\mathrm{}\times {10}^5}$  $=$ 5.263  $\times$  ${10}^{-6}{m}^3=5.263{cm}^3$

Q.13.6 Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0  ${\mathit{m}}^3$ at a temperature of 27 °C and 1 atm pressure.

Ans.13.6

Volume of the room, V = 25.0  $m^3$

Temperature of the room, T = 27  $℃$ = 300 K

Pressure of the room, P = 1 atm = 1  $\times$ 1.013  $\times {10}^5$ Pa

The ideal gas equation relating to pressure (P), volume (V) and absolute temperature (T) can be written as

PV =  $k_B$ NT, where  $k_B$ is Boltzmann constant = 1.38  $\times {10}^{-23}$  $m^{2}kg{s}^{-2}{K}^{-1}$

N is the number of air molecules in the room.

N =  $\frac{PV}{k_{B}T}$ =  $\frac{1\mathrm{}\times 1.013\mathrm{}\times {10}^5\times 25}{1.38\mathrm{}\times {10}^{-23}\times 300}$ = 6.117  $\times {10}^{26}$

Q.13.7 Estimate the average thermal energy of a helium atom at (i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million Kelvin (the typical core temperature in the case of a star).

Ans.13.7

At room temperature, T = 27  $℃$ = 300 K

$k_B$ is Boltzmann constant = 1.38  $\times {10}^{-23}$  $m^{2}kg{s}^{-2}{K}^{-1}$

Average thermal energy =  $\frac{3}{2}{k}_{B}T$ =  $\frac{3\times 1.38\mathrm{}\times {10}^{-23}\times 300}{2}$ = 6.21  $\times {10}^{-21}$ J

Hence, the average thermal energy of a helium atom at room temperature is 6.21  $\times {10}^{-21}$ J

On the surface of the Sun, T = 6000 K

Hence average thermal energy =  $\frac{3}{2}{k}_{B}T$ =  $\frac{3\times 1.38\mathrm{}\times {10}^{-23}\times 6000}{2}$ = 1.242  $\times {10}^{-19}$ J

Hence, the average thermal energy of a helium atom on the surface of the Sun is 1.242  $\times {10}^{-19}$ J

Inside the core of a star, T =  ${10}^7$ K

Hence average thermal energy =  $\frac{3}{2}{k}_{B}T$ =  $\frac{3\times 1.38\mathrm{}\times {10}^{-23}\times {10}^7\mathrm{}}{2}$ = 2.07  $\times {10}^{-16}$ J

Hence, the average thermal energy of a helium atom inside the core of a star is 2.07  $\times {10}^{-16}$ J

Q.13.8 Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms the largest?

Ans.13.8

According to Avogadro’s law, the three vessels will contain an equal number of the respective molecules. This number is equal to

Avogadro’s number, N = 6.023  $\times {10}^{23}$

The root mean square speed (  $V_{rms})$ of a gas of mass m and temperature T is given by the relation  $V_{rms}=\sqrt{\frac{3kT}{m}}$ . Where k is Boltzmann constant. For the given gases, k and T are constants. Hence  $V_{rms}$ depends only on the mass of the atoms  $V_{rms}\alpha \sqrt{\frac{1}{}}$ $$Therefore, the root mean square speed of the molecules in the three cases is not the same. Among Neon, Chlorine and Uranium hexafluoride, the mass of the neon is the smallest, so Neon will have highest root mean square speed among the given gases.

Q.13.9 At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C ? (atomic mass of Ar = 39.9 u, of He = 4.0 u).

Ans.13.9

Temperature of the helium atom,  $T_{He}$ = – 20 °C = 253 K and temperature of argon atom be =  $T_{Ar}$

Atomic mass of helium,  $M_{He}$ = 4.0 u

Atomic mass of Argon,  $M_{Ar}$ = 39.9 u

Let  ${V_{rms}}_{Ar}$ be the rms speed of Argon and  ${V_{rms}}_{He}$ be the rms speed of Helium

From the relation of  $V_{rms}=\sqrt{\frac{3kT}{M}}$ we get

rms speed of Argon,  ${V_{rms}}_{Ar}\mathrm{}=\sqrt{\frac{3k{T}_{Ar}}{M_{Ar}}}$

rms speed of Helium,  ${V_{rms}}_{He}\mathrm{}\mathrm{}=\sqrt{\frac{3k{T}_{He}}{M_{He}}}$

Since both the speeds are equal, we get

$\sqrt{\frac{3k{T}_{Ar}}{M_{Ar}}}$ =  $\sqrt{\frac{3k{T}_{He}}{M_{He}}}$ or  $\frac{T_{Ar}}{M_{Ar}}$ =  $\frac{T_{He}}{M_{He}}$ or  $T_{Ar}$ =  $\frac{T_{He}\times M_{Ar}}{M_{He}}$ =  $\frac{253\times 39.9}{4}$ = 2523.675 K = 2.523  $\times {10}^3$ K

Q.13.10 Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17  $\mathit{℃}$ . Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of  ${\mathit{N}}_2$ = 28.0 u).

Ans.13.10

Pressure inside the cylinder containing nitrogen, P = 2.0 atm = 2  $\times 1.013\times {10}^5$ Pa

Temperature inside the cylinder, T = 17  $℃=290K$

Radius of nitrogen molecule, r = 1.0 Å = 1  $\times {10}^{-10}$ m

Diameter of nitrogen molecule, d = 2  $\times {10}^{-10}$ m

Molecular mass of nitrogen molecule, M = 28 u = 28 g (assume) = 28  $\times {10}^{-3}$ kg

The root means square speed of nitrogen is given by the relation

R is the universal gas constant = 8.314 J/mole/K

Hence  $V_{rms}=\sqrt{\frac{3RT}{M}}$ = 508.26 m/s

The mean free path  $V_{rms}=\sqrt{\frac{3\times 8.314\times 290}{28\mathrm{}\times {10}^{-3}}}$ is given by

$\bar{l}$ where k = Boltzmann constant = 1.38  $l ̅=\frac{kT}{\surd 2\pi d^{2}P}$ kg-  $\times {10}^{-23}$

$m^2{s}^{-2}{K}^{-1}$ = 1.11  $\bar{l}=\frac{1.38\mathrm{}\times {10}^{-23}\times 290}{\surd 2\pi {(2\mathrm{}\times {10}^{-10})}^2\times 2\mathrm{}\times 1.013\times {10}^5}$ m

Collision frequency=  $\times {10}^{-7}$ =  $\frac{V_{rms}}{\bar{l}}$ = 4.57  $\frac{508.26}{1.11\mathrm{}\times {10}^{-7}}$ /s

Collision time, T =  ${10}^9$ =  $\frac{d}{V_{rms}}$ s = 3.93  $\frac{2\mathrm{}\times {10}^{-10}\mathrm{}}{508.26\mathrm{}}$ s

Time taken between successive collision T’ =  $\times {10}^{-13}$ =  $\frac{\bar{l}}{V_{rms}}$ s = 2.18  $\frac{1.11\mathrm{}\times {10}^{-7}}{508.26}$ s

$\times {10}^{-10}$ =  $\frac{\mathrm{T}\mathrm{’}}{T}$ = 555.71

Hence, the time taken between successive collisions is 556 times the time taken for a collision.

Q.13.11 A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?

Ans.13.11

Length of the narrow bore, L = 1 m = 100 cm

Length of the mercury thread, l = 76 cm

Length of the air column between mercury and the closed end,  $\frac{2.18\mathrm{}\times {10}^{-10}}{3.93\mathrm{}\times {10}^{-13}}$ = 15 cm

Since the bore is held vertically in air with the open end at the bottom, the mercury length that occupies the air space is 100 –(76 + 15) = 9 cm

Hence, total length of the air column = 15 + 9 = 24 cm

Let h cm of mercury flow out as a result of atmospheric pressure.

Length of the air column in the bore = 24 + h cm

Length of the mercury column = 76 – h cm

Initial pressure,  $l_o$ = 76 cm of mercury

Initial volume,  $P_1$ = 15  $V_1$

Final pressure,  ${cm}^3$ = 76 – (76 – h) = h cm of mercury

Final volume,  $P_2$ = (24 + h)  $V_2$

Since temperature remained constant throughout the process,

${cm}^3$ =  $P_1{V}_1$ or 76  $P_2{V}_2$ = h  $\times 15$

Or,  $\times (24+h)$ + 24h - 76  $h^2$ =0

h =  $\times 15$ =  $\frac{-24\pm \sqrt{{24}^2+4\times 76\times 15}}{2}$

So h = 23.84 cm or -47.84 cm

Since height cannot be negative, so h = 23.84 cm

23.84 cm of mercury will flow out from the bore and (76-23.84) 52.16 cm of mercury will remain in the bore. The length of the air column will be 24 + 23.84 = 47.84 cm

Q.13.12 From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm³ s^–1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³ s^–1. Identify the gas.

[Hint : Use Graham’s law of diffusion: R1/R2 = ( M₂ /M₁ )^1/2, where R1, R2 are diffusion rates of gases 1 and 2, and M1 and M2 their respective molecular masses. The law is a simple consequence of kinetic theory.]

Ans.13.12

Rate of diffusion of hydrogen,  $\frac{-24\pm 71.67}{2}$ = 28.7 cm3 s–1

Rate of diffusion of another gas,  $R_1$ = 7.2 cm3 s–1

According to Graham’s law of diffusion, we have:

$R_2$ =  $\frac{R_1}{R_2}$ , where  $\sqrt{\frac{M_2}{M_1}}$ = molecular mass of hydrogen = 2.02 g and  $M_1$ is the molecular mass of the unknown gas

$M_2$ = 2.02  $M_2=M_1\times ({\frac{R_1}{R_2}\mathrm{})}^2$ = 32.09 = Molecular mass of Oxygen

Hence, the unknown gas is Oxygen.

Q.13.13 A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres

n2 = n1 exp [ -mg (h2 – h1)/ kBT]

where n2, n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:

n2 = n1 exp [ -mg NA (  $\times ({\frac{28.7}{7.2}\mathrm{})}^2$ -  $\mathit{\rho }$ ) (h2 –h1)/ (  ${\mathit{\rho }}^{\mathit{\text{'}}}$ RT)]

where  $\mathit{\rho }$ is the density of the suspended particle, and  $\mathit{\rho }$ that of surrounding medium. [NA is Avogadro’s number, and R the universal gas constant.] [Hint : Use Archimedes principle to find the apparent weight of the suspended particle.]

Ans.13.13

According to law of atmospheres, we have

n2 = n1 exp [ -mg (h2 – h1)/ kBT] …..(i)

where  ${\mathit{\rho }}^{\mathit{\text{'}}}$ is the number of density at height  $n_1$ and  $h_1$ is the number of density at height  $n_2$

mg is the weight of the particle suspended in the gas column

Density of the medium =  $h_2$

Density of the suspended particle =  $\rho$

Mass of one suspended particle = m’

Mass of medium displaced = m

Volume of the suspended particle = V

According to Archimedes’s principle for a particle suspended in a liquid column, the effective weight of the suspended particle is given as:

Weight of the medium displaced – weight of the suspended particle = mg – m’g

= mg- V  ${\rho }^{\text{'}}$ = mg – (  ${\rho }^{\text{'}}g$ )  $\frac{m}{\rho }$

= mg(1-  ${\rho }^{\text{'}}g$ ) ……(ii)

Gas constant R =  $\frac{{\rho }^{\text{'}}}{\rho }$ N

$k_B$ …(iii)

Substituting from (ii) and (iii) in (i), we get

n2 = n1 exp [ -mg (h2 – h1)/ kBT]

= n1 exp  $k_B=\frac{R}{N}$

= n1 exp  $\left[-mg\left(-\frac{{\rho }^{\text{'}}}{\rho }\right)(h2\mathrm{}–\mathrm{}h1)\frac{N}{RT}\right]$

Q.13.14 Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:

[Hint : Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å].

Ans.13.14

Let the atomic mass of a substance be = M and the density of the substance be =  $\left[-mg\left(\rho -{\rho }^{\text{'}}\right)(h2\mathrm{}–\mathrm{}h1)\frac{N}{RT\rho }\right]$

Avogadro’s number, N = 6.023  ${10}^3\mathit{K}\mathit{g}{\mathit{m}}^{-3})$

Volume of N number of molecules =  $\rho$ ……..(i)

Volume of one mole of a substance =  $\times {10}^{23}$ ……(ii)

Equating (i) and (ii), we get

r =  $\frac{4}{3}\pi r^{3}N$

For diamond, M = 12.01  $\frac{M}{\rho }$ kg,  $\frac{4}{3}\pi r^{3}N=\frac{M}{\rho }$ = 2.22  $\sqrt[3]{\frac{3M}{4\pi \rho N}}$ kg/  $\times {10}^{-3}$

$\rho$ =  $\times {10}^3$ = 1.29 Å

For gold, M = 197  $m^3$ kg,  $r_{diamond}$ = 19.32  $\sqrt[3]{\frac{3\times 12.01\mathrm{}\times {10}^{-3}}{4\times \pi \times 2.22\mathrm{}\times {10}^3\times 6.023\mathrm{}\times {10}^{23}}}$ kg/  $\times {10}^{-3}$

$\rho$ =  $\times {10}^3$ = 1.59 Å

For liquid nitrogen, M = 14.01  $m^3$ kg,  $r_{gold}$ = 1.0  $\sqrt[3]{\frac{3\times 197\mathrm{}\times {10}^{-3}}{4\times \pi \times 19.32\mathrm{}\times {10}^3\times 6.023\mathrm{}\times {10}^{23}}}$ kg/  $\times {10}^{-3}$

$\rho$ =  $\times {10}^3$ = 1.77 Å

For lithium, M = 6.94  $m^3$ kg,  $r_{nitrogen}$ = 0.53  $\sqrt[3]{\frac{3\times 14.01\mathrm{}\times {10}^{-3}}{4\times \pi \times 1.0\mathrm{}\times {10}^3\times 6.023\mathrm{}\times {10}^{23}}}$ kg/  $\times {10}^{-3}$

$\rho$ =  $\times {10}^3$ = 1.73 Å

For fluorine, M = 19  $m^3$ kg,  $r_{lithium}$ = 1.14  $\sqrt[3]{\frac{3\times 6.94\mathrm{}\times {10}^{-3}}{4\times \pi \times 0.53\mathrm{}\times {10}^3\times 6.023\mathrm{}\times {10}^{23}}}$ kg/  $\times {10}^{-3}$

$\rho$ =  $\times {10}^3$ = 1.88 Å