Problem 32 - Chemistry The Central Science

Question

Sodium metal requires a photon with a minimum energy of \(4.41 \times 10^{-19} \mathrm{~J}\) to emit electrons. (a) What is the minimum frequency of light necessary to emit electrons from sodium via the photoelectric effect? (b) What is the wavelength of this light? (c) If sodium is irradiated with light of \(405 \mathrm{nm},\) what is the maximum possible kinetic energy of the emitted electrons? (d) What is the maximum number of electrons that can be freed by a burst of light whose total energy is \(1.00 \mu \mathrm{J}\) ?

Step-by-Step Solution

Answer

The minimum frequency of light necessary to emit electrons from sodium via the photoelectric effect is \(6.65 \times 10^{14} Hz\). The wavelength of this light is \(4.51 \times 10^{-7} m\) or \(451 nm\). The maximum possible kinetic energy of the emitted electrons when sodium is irradiated with light of \(405 nm\) is \(1.98 \times 10^{-19} J\). The maximum number of electrons that can be freed by a burst of light with a total energy of \(1.00 \mu J\) is approximately \(2.27 \times 10^{12}\).

Step 1: To find the minimum frequency required to emit electrons from sodium via the photoelectric effect, we can use the formula: \( E = h \nu \) Rearrange the formula to find the frequency (\(\nu\)): \(\nu = E / h \) Substitute the given energy and Planck's constant: \(\nu = (4.41 \times 10^{-19} J) / (6.63 \times 10^{-34} Js) \) Perform the calculation: \(\nu = 6.65 \times 10^{14} Hz\) The minimum frequency of light necessary to emit electrons from sodium via the photoelectric effect is \(6.65 \times 10^{14} Hz\). #b) Find the Wavelength of Light#

To find the wavelength of the light, we can use the formula: \(\nu = c / \lambda \) Rearrange the formula to find wavelength (\(\lambda\)): \(\lambda = c / \nu \) Substitute the given speed of light and calculated frequency: \(\lambda = (3.0 \times 10^8 m/s) / (6.65 \times 10^{14} Hz) \) Perform the calculation: \(\lambda = 4.51 \times 10^{-7} m \) The wavelength of this light is \(4.51 \times 10^{-7} m\) or \(451 nm\). #c) Calculate Maximum Kinetic Energy of Emitted Electrons#

Step 2: We will use the following formula to calculate the maximum possible kinetic energy of emitted electrons: \(K_e = hf - W\) We can replace frequency (\(\nu\)) with speed of light/wavelength, to get: \(K_e = h (c/\lambda) - W\) Substitute the given wavelength, Planck's constant, speed of light and work function: \(K_e = (6.63 \times 10^{-34} Js) ((3.0 \times 10^8 m/s)/(405 \times 10^{-9} m)) - 4.41 \times 10^{-19} J \) Perform the calculation: \(K_e = 1.98 \times 10^{-19} J\) The maximum possible kinetic energy of the emitted electrons is \(1.98 \times 10^{-19} J\). #d) Find the Maximum Number of Electrons That Can Be Freed#

To find the maximum number of electrons that can be freed by a burst of light whose total energy is 1.00 μJ, we can use the following formula: \(N = E_{total} / E_{per \, electron}\) where: \(N\) = maximum number of electrons \(E_{total}\) = total energy of the burst of light \(E_{per \, electron}\) = energy required to free one electron Substitute the given total energy and calculated energy per electron: \(N = (1.00 \times 10^{-6} J) / (4.41 \times 10^{-19} J)\) Perform the calculation: \(N = 2.27 \times 10^{12}\) The maximum number of electrons that can be freed by a burst of light with a total energy of 1.00 μJ is approximately \(2.27 \times 10^{12}\).

Key Concepts

Photon Energy

Photon energy is the energy carried by a single photon, which is the basic unit of light and all other forms of electromagnetic radiation. It plays a fundamental role in understanding the photoelectric effect, as it determines whether an electron can be ejected from a material. The energy of a photon depends on its frequency and is given by the formula:

• \( E = h u \)

where:

• \( E \) is the energy of the photon,
• \( h \) is Planck's constant \((6.63 \times 10^{-34} \text{ Js})\),
• \( u \) is the frequency of the photon.

This means that the higher the frequency of a photon, the greater its energy. In practical terms, to cause an electron to be emitted from a metal surface, the photon's energy must be at least equal to the energy required to remove an electron from the surface, which is known as the work function of the material.

Frequency and Wavelength

Frequency and wavelength are intimately connected in the behavior of light and other forms of electromagnetic radiation. Frequency \(u\) refers to the number of wave cycles that pass a given point per second and is measured in hertz (Hz). Wavelength \(\lambda\) is the distance between successive peaks of a wave and is typically measured in meters.These two properties are inversely related through the speed of light \(c\), according to the equation:

• \( u = \frac{c}{\lambda} \)

where:

• \( c \) is the speed of light \((3.0 \times 10^{8} \text{ m/s})\).

As frequency increases, wavelength decreases, and vice versa. To find the wavelength of light knowing its frequency, you can rearrange the equation to:

• \( \lambda = \frac{c}{u} \)

Understanding the relationship between frequency and wavelength is essential for analyzing how different colors of light, which correspond to different frequencies and wavelengths, interact with various materials.

Kinetic Energy of Electrons

The kinetic energy of electrons ejected through the photoelectric effect is determined by the energy balance between the incoming photon and the work function of the material. The excess energy, after overcoming the electron's binding energy, translates into the kinetic energy of the electron. This is described by the equation:

• \( K_e = hu - W \)

where:

• \( K_e \) is the kinetic energy of the emitted electron,
• \( hu \) is the energy of the incoming photon,
• \( W \) is the work function of the material.

If we know the wavelength of the incoming light, we can substitute the relationship between frequency and wavelength to get:

• \( K_e = h\left(\frac{c}{\lambda}\right) - W \)

In practice, if the energy of the photon is just enough to overcome the work function, the electron will have zero kinetic energy. Additional energy from the photon will result in the electron being ejected with some kinetic energy, which can be calculated using the above formulas.