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3 months ago

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The mean free path of a helium atom in helium gas at standard temperature and pressure is 0.2 um.What is the radius of the heliu

m atom in nanometers?

1 answer:

Yuliya22 [3.3K]3 months ago

5 0

Answer: $0.10233nm$

Explanation:

The mean free path of an atom can be calculated using the following equation:

$\lambda=\frac{RT}{\sqrt{2} \pi d^{2}N_{A}P}$ (1)

Where:

$\lambda=0.2\mu m=0.2(10)^{-6}m$

$R=8.3145J/mol.K$ is referred to as the Universal gas constant

$T=0\°C=273.115K$ represents the absolute standard temperature

$d$ denotes the diameter of helium atoms

$N_{A}=6.0221(10)^{23}/mol$ symbolizes Avogadro's number

$P=1atm=101.3kPa=101.3(10)^{3}Pa=101.3(10)^{3}J/m^{3}$ indicates absolute standard pressure

<pFrom this, we can solve for $d$ using (1), aiming to determine the radius $r$ of the helium atom:

$d=\sqrt{\frac{RT}{\sqrt{2}\pi\lambda N_{A}P}}$ (2)

$d=\sqrt{\frac{(8.3145J/mol.K)(273.115K)}{\sqrt{2}\pi(0.2(10)^{-6}m)(6.0221(10)^{23}/mol)(101.3(10)^{3}J/m^{3})}}$ (3)

$d=2.0467(10)^{-10}m$ (4)

If the radius equals half of that diameter:

$r=\frac{d}{2}$ (5)

Eventually:

$r=\frac{2.0467(10)^{-10}m}{2}$ (6)

$r=1.0233(10)^{-10}m$ (7)

Nonetheless, we were tasked with finding this radius in nanometers. Knowing $1nm=(10)^{-9}m$ :

$r=1.0233(10)^{-10}m.\frac{1nm}{(10)^{-9}m}=0.10233nm$ (8)

Ultimately:

$r=0.10233nm$ Represents the radius of the helium atom in nanometers.

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A proton (mass = 1.67 10–27 kg, charge = 1.60 10–19 C) moves from point A to point B under the influence of an electrostatic for

serg [3582]

Answer:

20.353125 V

Explanation:

m = Mass of proton = $1.67\times 10^{-27}\ kg$

q = Charge of proton = $1.6\times 10^{-19}\ C$

$v_A$ = Velocity of proton at point A = 50 km/s

$v_B$ = Velocity of proton at point B = 80 km/s

The relationship derived from energy conservation is as follows:

$\dfrac{1}{2}m(v_B^2-v_A^2)=q(V_B-V_A)\\\Rightarrow V_B-V_A=\dfrac{1}{2q}m(v_B^2-v_A^2)\\\Rightarrow V_B-V_A=\dfrac{1}{2\times 1.6\times 10^{-19}}\times 1.67\times 10^{-27}(80000^2-50000^2)\\\Rightarrow V_B-V_A=20.353125\ V$

The determined potential difference is 20.353125 V

3 0

3 months ago

It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you

Maru [3345]

Answer:

0.0984

Explanation:

The first diagram below illustrates a free body diagram that will aid in resolving this problem.

According to the diagram, the force's horizontal component can be expressed as:

$F_X = F_{cos \ \theta}$

Substituting 42° for θ and 87.0° for $F$

$F_X =87.0 \ N \ *cos \ 42 ^\circ$

$F_X =64.65 \ N$

Meanwhile, the vertical component is:

$F_Y = Fsin \ \theta$

Again substituting 42° for θ and 87.0° for $F$

$F_Y =87.0 \ N \ *sin \ 42 ^\circ$

$F_Y =58.21 \ N$

In resolving the vector, let A denote the components in mutually perpendicular directions.

The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ

The frictional force can be expressed as:

$f = \mu \ N$

Where;

$\mu$ is the coefficient of friction

N = the normal force

Also, the normal reaction (N) is calculated as mg - F sin θ

Substituting $F_Y \ for \ F_{sin \ \theta}$ . Normal reaction becomes:

$N = mg \ - \ F_Y$

By balancing the forces, the horizontal component of the force equals the frictional force.

The horizontal component is described as follows:

$F_X = \mu \ ( mg - \ F_Y)$

Rearranging the equation above to isolate $\mu$ leads to:

$\mu \ = \ \frac{F_X}{mg - F_Y}$

Substituting in the following values:

$F_X \ = \ 64.65 \ N$

m = 73 kg

g = 9.8 m/s²

$F_Y = \ 58.21 N$

Thus:

$\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}$

$\mu = 0.0984$

Therefore, the coefficient of friction is = 0.0984

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3 months ago

In a demonstration, a 4.00 cm2 square coil with 10 000 turns enters a larger square region with a uniform 1.50 T magnetic field

serg [3582]

Explanation:

It is stated that,

Area of square coil, $A=4\ cm^2=0.0004\ m^2$

Length of one side of the square, L = 0.02 m

Number of coils, N = 10000

Consistent magnetic field, B = 1.5 T

Velocity, v = 100 m/s

An electromotive force is produced in the coil calculated as:

$E=NBLv$

$E=10000\times 1.5\times 0.02\times 100$

E = 30000 V

Breakdown voltage of air, $V=4000\ V/cm=400000\ V/m$

Let d be the distance between the ends of the coil wires that can still produce a spark. Therefore,

Electric field, $E'=\dfrac{V}{d}$

$\dfrac{30000}{d}=400000$

d = 0.075 m

Thus, this forms the final answer.

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2 months ago