Question

To eject electrons from the surface of potassium metal requires a minimum energy of \(3.68 \times 10^{-19} \mathrm{~J}\). When \(600 .-\mathrm{nm}\) photons shine on a potassium surface, will they cause the photoelectric effect? Explain.

Short answer

The 600-nm photons do not cause the photoelectric effect in potassium.

Step-by-step solution

Understand the Problem

First, we need to determine if the incoming photons have enough energy to overcome the minimum energy required to eject electrons (the work function) from potassium. The work function is given as \(3.68 \times 10^{-19} \text{ J}\). We need to calculate the energy of the 600-nm photons and compare it with this value.

Convert Wavelength to Energy

Use the formula for the energy of a photon: \[ E = \frac{hc}{\lambda} \] where \( E \) is the energy of the photon, \( h = 6.626 \times 10^{-34} \text{ J s} \) is the Planck's constant, \( c = 3.00 \times 10^8 \text{ m/s} \) is the speed of light, and \( \lambda = 600 \times 10^{-9} \text{ m} \) is the wavelength of the photon.

Apply the Formula

Substitute the given values into the formula to find the energy of the 600-nm photon:\[ E = \frac{(6.626 \times 10^{-34} \text{ J s})(3.00 \times 10^8 \text{ m/s})}{600 \times 10^{-9} \text{ m}} \] Calculate \( E \).

Calculate Energy of 600-nm Photon

Calculating the expression:\[ E = \frac{1.9878 \times 10^{-25} \text{ J m}}{600 \times 10^{-9} \text{ m}} = 3.31 \times 10^{-19} \text{ J} \] This energy represents the energy of a single photon with 600-nm wavelength.

Compare Photon Energy to Work Function

Now, compare the calculated photon energy \(3.31 \times 10^{-19} \text{ J}\) to the minimum energy required \(3.68 \times 10^{-19} \text{ J}\). Since the photon energy is less than the work function, the photons do not have sufficient energy to eject electrons from the potassium surface.

Key concepts

Work Function

The work function is fundamental in understanding the photoelectric effect, dictating how much energy is needed to remove an electron from a metal surface. Think of it as the threshold that photons must meet to free an electron from the grip of a metal atom.
In this context, potassium metal requires a work function of \(3.68 \times 10^{-19} \text{ J}\). This means any photon aimed at the surface must possess at least this energy to succeed in ejecting an electron.
Work function values vary among different metals. They depend on the metal's atomic structure, electron bonding, and surface conditions. For potassium, known for its low work function compared to other metals, this is an important quality.
Understanding a metal's work function helps in designing devices like solar cells and photo detectors, where maximizing electron emission is vital.

Photon Energy Calculation

To ascertain whether a photon can eject an electron, we calculate its energy using its wavelength. The energy of a photon is linked to its wavelength by the formula:

• \(E = \frac{hc}{\lambda}\)

Here, \(E\) is the energy, \(h\) (Planck's constant) is \(6.626 \times 10^{-34} \text{ J s}\), \(c\) (speed of light) is \(3.00 \times 10^8 \text{ m/s}\), and \(\lambda\) is the wavelength.
Consider photons with a wavelength of 600-nm. Substitute these values into the formula:

• \(E = \frac{(6.626 \times 10^{-34} \text{ J s})(3.00 \times 10^8 \text{ m/s})}{600 \times 10^{-9} \text{ m}}\)
• \(E = 3.31 \times 10^{-19} \text{ J}\)

This result \(3.31 \times 10^{-19} \text{ J}\) shows whether the photons have enough energy when analyzed against the work function.

Potassium Metal

Potassium is a soft, easily oxidizing metal, making it unique in photoelectric experiments. Its low work function makes it an excellent candidate for such effects and demonstrations.
Being an alkali metal, potassium's electrons are not tightly bound to the nucleus. This characteristic is evident in its relatively low minimum energy requirement for electron emission compared to other metals.
In many practical applications, such as photoelectric sensors, the choice of metal affects the efficiency. Potassium's properties explain why it’s favored in applications needing efficient electron emission under light exposure.
However, understanding its limitations is vital. In the given exercise, photons with 600-nm wavelength are not energetic enough to surpass potassium's work function of \(3.68 \times 10^{-19} \text{ J}\). It highlights the relationship between material choice and photon energy needed for specific outputs.