When a yellow light shines on a photosensitive metal, photoelectrons are emitted. As the intensity of the light is decreased, the number of photoelectrons emitted per second
decreases
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As the intensity of the light is decreased, the number of photoelectrons emitted per second also decreases. This is due to the fact that the photoelectric effect relies on the energy of the photons (particles of light) to liberate electrons from the metal surface.
When photons with sufficient energy strike the photosensitive metal, they transfer their energy to the electrons in the metal, which allows the electrons to overcome their binding energy and be emitted as photoelectrons. The intensity of the light is directly proportional to the number of photons striking the metal per unit time.
Therefore, if the intensity of the light is decreased, fewer photons will strike the metal, resulting in a lower rate of photoelectron emission. This decrease in intensity typically leads to a decrease in the number of photoelectrons emitted per second.
To understand the relationship between the intensity of the yellow light and the number of photoelectrons emitted per second, we need to explore the phenomenon known as the photoelectric effect.
The photoelectric effect refers to the emission or release of free electrons (photoelectrons) from a material when it is exposed to light. This effect can be observed in photosensitive metals, which are materials that have the ability to release electrons when illuminated.
When a yellow light shines on a photosensitive metal, its photons (particles of light) interact with the atoms or molecules on the metal's surface. These interactions can lead to the ejection of electrons from the metal, creating photoelectrons.
The intensity of light refers to the amount of energy carried by each photon or the number of photons incident on the metal's surface per unit time. In the context of the photoelectric effect, the intensity of light determines the number of photoelectrons emitted per second.
According to the laws of the photoelectric effect, several factors influence the emission of photoelectrons:
1. Intensity of Light: The number of photons incident on the metal's surface per unit time determines the number of photoelectrons emitted. As the intensity of yellow light increases, more photons interact with the metal, leading to a higher number of photoelectrons being emitted per second.
2. Frequency of Light: In addition to intensity, the frequency (or color) of light also plays a crucial role. Different metals have different energy thresholds for the photoelectric effect. If the incoming photons do not possess sufficient energy (frequency) to overcome the metal's binding energy, no photoelectrons will be emitted, regardless of the intensity.
However, if the incoming yellow light has a frequency that exceeds the threshold, each photon can contribute enough energy to liberate an electron from the metal's surface. In this case, increasing the intensity of the yellow light will result in more photoelectrons being emitted per second.
To quantitatively determine the relationship between the intensity of yellow light and the number of photoelectrons emitted per second, experimental data needs to be collected. By changing the intensity of the yellow light while keeping other factors constant (such as frequency and distance), one can measure the corresponding number of photoelectrons emitted per second. Analyzing these data points will allow for the establishment of a relationship or pattern between the two variables.