Question

Consider an electron in a hydrogen atom, which is \(0.050 \mathrm{nm}\) from the proton in the nucleus. a) What electric field does the electron experience? b) In order to produce an electric field whose root-meansquare magnitude is the same as that of the field in part (a), what intensity must a laser light have?

Short answer

a) The electric field experienced by the electron in the hydrogen atom due to the proton in the nucleus is approximately \(4.58 * 10^{11} \mathrm{N/C}\). b) The intensity of a laser light needed to produce a root-mean-square magnitude of the electric field calculated in part (a) is approximately \(2.83 * 10^9 \mathrm{W/m^2}\).

Step-by-step solution

Find the electric field experienced by the electron

To find the electric field experienced by the electron, we'll use the formula for Coulomb's law, imagining that there's an electron at \(0.050 \mathrm{nm}\) from the proton in the nucleus: \( E = \frac{k * q_1 * q_2}{r^2} \) Where: - E is the electric field, - k is the electrostatic constant (\(8.99 * 10^9 \mathrm{Nm^2C^{-2}}\)), - \(q_1\) and \(q_2\) are the charges of the proton and electron, respectively (both have the same magnitude of charge \(e\): \(1.6 * 10^{-19}\ \mathrm{C}\)), - r is the distance between the proton and electron (\(0.050\ \mathrm{nm}\)). Plugging in the values: \( E = \frac{(8.99 * 10^9 \mathrm{Nm^2C^{-2}}) * (1.6 * 10^{-19}\ \mathrm{C})^2}{(0.050 * 10^{-9}\ \mathrm{m})^2} \) After calculating, we get: \( E \approx 4.58 * 10^{11}\ \mathrm{N/C} \)

Relate the electric field with the intensity of the laser light

We will now find the relationship between the electric field and the intensity of a laser light. In a plane electromagnetic wave, we know that: \( S = \frac{1}{2} * {E_{rms}}^2 * \epsilon_0 * c \) Where: - S is the power per unit area (intensity), - \(E_{rms}\) is the root-mean-square electric field, - \(\epsilon_0\) is the vacuum permittivity (\(\approx 8.85 * 10^{-12} \mathrm{C^2/Nm^2}\)), - c is the speed of light in vacuum (\(\approx 3 * 10^8 \mathrm{m/s}\)). Since we want the intensity to have the same root-mean-square magnitude as the electric field calculated in part (a), we can rewrite the above equation as: \( S = \frac{1}{2} * E^2 * \epsilon_0 * c \)

Calculate the intensity of the laser light to produce the given root-mean-square electric field magnitude

Plugging in the values from step 1 for the electric field, and the values for \(\epsilon_0\) and c: \( S = \frac{1}{2} * ({4.58 * 10^{11}\ \mathrm{N/C}})^2 * (8.85 * 10^{-12}\ \mathrm{C^2/Nm^2}) * (3 * 10^8\ \mathrm{m/s}) \) Calculating this expression, we get: \( S \approx 2.83 * 10^9 \mathrm{W/m^2} \) So in order to produce an electric field with the same root-mean-square magnitude as that in the hydrogen atom, the laser light must have an intensity of approximately \(2.83 * 10^9 \mathrm{W/m^2}\).

Key concepts

Coulomb's Law

Coulomb’s Law is crucial for understanding the electric field between two charged particles, like an electron and a proton. This law gives us the electric force between them. The formula is:

• \( E = \frac{k * q_1 * q_2}{r^2} \),
• where:
  • \( E \) is the electric field strength,
  • \( k \) is Coulomb’s constant, approximately \( 8.99 \times 10^9 \ \text{Nm}^2/\text{C}^2 \),
  • \( q_1 \) and \( q_2 \) are the charges, both \( 1.6 \times 10^{-19} \ \text{C} \) for a proton and an electron,
  • \( r \) is the distance between the particles.

To find an electric field, imagine shrinking this distance to the atomic level, like in a hydrogen atom. Using Coulomb’s Law for an electron at \( 0.050 \ \text{nm} \) from the nucleus, you would substitute the values into the formula, resulting in a powerful electric field of about \( 4.58 \times 10^{11} \ \text{N/C} \). This is the intense electric environment experienced by particles within an atom.

Electromagnetic Wave Intensity

The intensity of an electromagnetic wave is its power per unit area, showing how much energy passes through a given surface. In this context, intensity helps to understand how light and other waves interact with their surroundings. The formula for intensity \( S \) is related to the electric field by:

• \( S = \frac{1}{2} * E_{rms}^2 * \epsilon_0 * c \).
• Here:
  • \( E_{rms} \) represents the root-mean-square of the electric field, reflecting its average value over time,
  • \( \epsilon_0 \) is the vacuum permittivity, around \( 8.85 \times 10^{-12} \ \text{C}^2/\text{Nm}^2 \),
  • \( c \) is the speed of light, approximately \( 3 \times 10^8 \ \text{m/s} \).

In scenarios like laser applications, this relationship tells us how the electric field from a laser must be matched by its intensity. For instance, achieving an intensity of \( 2.83 \times 10^9 \ \text{W/m}^2 \) would complement the electric field strength encountered inside a hydrogen atom. This is especially useful in fields such as spectroscopy and quantum mechanics.

Root-Mean-Square Electric Field

The root-mean-square (RMS) electric field is a measure used to describe the strength of a varying electric field, often in AC circuits or electromagnetic waves. It provides an average value that reflects the overall effect and intensity of the field.

The RMS value is particularly important when dealing with waves, as it accounts for fluctuations over time. When determining the intensity of laser light with the same RMS electric field as that in an atom, the RMS formula helps equate the intensity and field magnitude:

• \( E_{rms} = \sqrt{\text{Average of } E^2} \).

By calculating this value, we can ensure that the laser or any other wave source replicates the atomic environment's conditions. This is a vital concept in both theoretical and applied sciences, helping transition calculations from microscopic atomic levels to observable scenarios.