An electron is orbiting a nucleus which has a charge of 16e, under the action of the Coulomb force at a radius of 1.97 × 10^-10 m. Calculate the angular velocity of the electron, i... An electron is orbiting a nucleus which has a charge of 16e, under the action of the Coulomb force at a radius of 1.97 × 10^-10 m. Calculate the angular velocity of the electron, in radians per second.
Understand the Problem
The question is asking us to calculate the angular velocity of an electron orbiting a nucleus with a specific charge, under the influence of the Coulomb force, given the radius of the orbit. To solve this, we will apply the formula for centripetal force and equate it to the Coulomb force acting on the electron.
Answer
$$ \omega = \sqrt{\frac{8.99 \times 10^9 \cdot (1.602 \times 10^{-19})^2}{9.11 \times 10^{-31} \cdot r^3}} $$
Answer for screen readers
$$ \omega = \sqrt{\frac{8.99 \times 10^9 \cdot (1.602 \times 10^{-19})^2}{9.11 \times 10^{-31} \cdot r^3}} $$
Steps to Solve
- Identify the Forces Acting on the Electron
The electron experiences two main forces: the Coulomb force due to the nucleus and the centripetal force required for circular motion. We'll use the equation for the Coulomb force:
$$ F_c = \frac{k \cdot |q_1 \cdot q_2|}{r^2} $$
Here, $k$ is Coulomb's constant, $q_1$ is the charge of the nucleus, $q_2$ is the charge of the electron (which is negative), and $r$ is the radius of the orbit.
- Centripetal Force Expression
The centripetal force $F_{centripetal}$ needed to keep the electron moving in a circular path is given by:
$$ F_{centripetal} = m \cdot \omega^2 \cdot r $$
where $m$ is the mass of the electron, $\omega$ is the angular velocity, and $r$ is the radius of the orbit.
- Equate the Forces
Since the centripetal force is provided by the Coulomb force, we set the two forces equal to each other:
$$ \frac{k \cdot |q_1 \cdot q_2|}{r^2} = m \cdot \omega^2 \cdot r $$
- Rearrange to Solve for Angular Velocity
Rearranging the equation to find the angular velocity $\omega$, we get:
$$ \omega^2 = \frac{k \cdot |q_1 \cdot q_2|}{m \cdot r^3} $$
Now, taking the square root we find:
$$ \omega = \sqrt{\frac{k \cdot |q_1 \cdot q_2|}{m \cdot r^3}} $$
- Substitute the Values
Plug in known values:
- $k \approx 8.99 \times 10^9 , \text{N m}^2/\text{C}^2$
- $q_1$ (nucleus for a hydrogen atom) $= 1.602 \times 10^{-19} , \text{C}$
- $q_2$ (electron charge) $= -1.602 \times 10^{-19} , \text{C}$
- Mass of electron $m = 9.11 \times 10^{-31} , \text{kg}$
- Radius $r = \text{(given radius in meters)}$
Substituting these values into the $\omega$ equation will yield angular velocity.
$$ \omega = \sqrt{\frac{8.99 \times 10^9 \cdot (1.602 \times 10^{-19})^2}{9.11 \times 10^{-31} \cdot r^3}} $$
More Information
The angular velocity $\omega$ calculated using this approach shows how fast the electron is orbiting the nucleus under the influence of the Coulomb force. This concept is fundamental in atomic physics and helps explain the behavior of electrons in different orbitals.
Tips
- Incorrectly substituting signs for the charges; remember that we are concerned about magnitudes in the Coulomb force equation.
- Forgetting to square the radius when equating the centripetal and Coulomb forces.
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