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Fiesta28 [93]
3 years ago
9

g The potential energy of a pair of hydrogen atoms separated by a large distance x is given by U(x)=−C6/x6, where C6 is a positi

ve constant. Part A What is the force that one atom exerts on the other? Express your answer in terms of C6 and x. Fx = nothing Request Answer Part B Is this force attractive or repulsive? Is this force attractive or repulsive? attractive repulsive
Physics
1 answer:
Arisa [49]3 years ago
5 0

Answer:

F_x = -\frac{6 C_6}{2^7}

Attractive

Explanation:

Data provided in the question

The potential energy of a pair of hydrogen atoms given by \frac{C_6}{X_6}

Based on the given information, the force that one atom exerts on the other is

Potential energy μ = \frac{C_6}{X_6}

Force exerted by one atom upon another

F_x = \frac{\partial U}{\partial X} = \frac{\partial}{\partial X}  (-\frac{C_6}{X^6})

or

F_x = \frac{\partial}{\partial X}  (\frac{C_6}{X^6})

or

F_x = -\frac{6 C_6}{2^7}

As we can see that the C_6 comes in positive and constant which represents that the force is negative that means the force is attractive in nature

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A) v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

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Mass of first particle; m1 = 5.03 × 10^(-27) kg

Speed of first particle in y-direction; v1 = (6 × 10^(6) m/s) j^

Mass of second particle; m2 = 8.47 × 10^(-27) kg

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Now, we don't have the mass of the third particle but since we are told the unstable atomic nucleus disintegrates into 3 particles, thus;

M = m1 + m2 + m3

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MV = (m1 × v1) + (m2 × v2) + (m3 × v3)

Now, the unstable atomic nucleus was at rest before disintegration, thus V = 0 m/s.

Thus, we now have;

0 = (m1 × v1) + (m2 × v2) + (m3 × v3)

We want to find the velocity of the third particle v3. Let's make it the subject of the formula;

v3 = [(m1 × v1) + (m2 × v2)]/(-m3)

Plugging in the relevant values, we have;

v3 = [(5.03 × 10^(-27) × 6 × 10^(6))j^ + (8.47 × 10^(-27) × 4 × 10^(6))i^]/(-4.8 × 10^(-27))

v3 = [(30.18 × 10^(-21))j^ + (33.88 × 10^(-21))i^]/(-4.8 × 10^(-27))

v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) Formula for kinetic energy is;

K = ½mv²

Now,total kinetic energy is;

K_total = K1 + K2 + K3

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K3 = 214.78 × 10^(-15) J

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K_total = 373.08 × 10^(-15) J

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