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Post a LessonAnswered on 06 Apr Learn CBSE/Class 12/Science/Physics/Unit 7-Dual Nature of Matter
Sadika
In the photon picture of light, the intensity of radiation refers to the rate at which photons carry energy per unit area perpendicular to the direction of propagation. It represents the amount of radiant energy per unit time and per unit area.
Mathematically, the intensity of radiation (II) is defined as:
I=PAI=AP
Where:
The SI unit of intensity of radiation is watts per square meter (W/m2W/m2). It represents the power per unit area and is commonly used to quantify the brightness or strength of radiation.
Answered on 06 Apr Learn CBSE/Class 12/Science/Physics/Unit 7-Dual Nature of Matter
Sadika
To determine which of the two photosensitive metals, A or B, has a higher value of the work function, we need to analyze their respective graphs showing the variation of stopping potential (VsVs) with frequency (ff) of incident radiation.
The work function (WW) of a metal is the minimum energy required to remove an electron from the surface of the metal and is directly related to the threshold frequency (f0f0) by the equation:
W=hf0W=hf0
where:
From the graph, the stopping potential (VsVs) is directly proportional to the frequency (ff) of the incident radiation and is given by the equation:
Vs=hfe−WeVs=ehf−eW
where:
By analyzing the graph, we can observe the following:
Since the work function is proportional to the intercept, the metal with the higher stopping potential (i.e., the metal with the higher intercept on the VsVs axis) has a higher work function.
Therefore, in the given graph, the metal associated with the higher stopping potential (further above the x-axis) has the higher value of the work function.
Answered on 06 Apr Learn CBSE/Class 12/Science/Physics/Unit 7-Dual Nature of Matter
Sadika
Without access to the graph you mentioned, I can't directly analyze the data to determine which of the two photosensitive metals, A or B, has a higher threshold frequency. However, I can provide a general explanation based on the concept of threshold frequency.
The threshold frequency (f0f0) is the minimum frequency of incident radiation required to eject electrons from the surface of a photosensitive material. Metals with higher work functions typically have higher threshold frequencies.
Here's why:
Work Function Relationship: The work function (WW) of a material is directly proportional to its threshold frequency (f0f0) according to the equation W=hf0W=hf0, where hh is Planck's constant. Therefore, a higher work function corresponds to a higher threshold frequency.
Ejection of Electrons: For electrons to be emitted from the surface of a metal when exposed to light, the energy of the incident photons must be greater than or equal to the work function (hf0hf0). If the frequency of the incident radiation is lower than the threshold frequency (f0f0), the photons do not possess enough energy to overcome the work function, and no electrons are emitted.
Comparison: Based on the concept above, the metal with the higher stopping potential (associated with the higher work function) likely has the higher threshold frequency. This is because it requires higher-energy photons (higher frequency) to eject electrons from its surface compared to the metal with the lower stopping potential.
Therefore, without the specific data from the graph, we can infer that the metal associated with the higher stopping potential (likely metal B) has the higher threshold frequency due to its higher work function.
Answered on 06 Apr Learn CBSE/Class 12/Science/Physics/Unit 7-Dual Nature of Matter
Sadika
In the photoelectric effect, the photoelectric current is the flow of electrons emitted from a photosensitive surface when it is illuminated by light. The intensity of monochromatic radiation refers to the amount of light energy incident per unit area per unit time. When the intensity of monochromatic radiation incident on a photosensitive surface is increased, several factors contribute to an increase in the photoelectric current:
Greater Number of Photons: Increasing the intensity of monochromatic radiation means that there are more photons striking the photosensitive surface per unit time. Each photon has a certain amount of energy associated with it. With a higher intensity, there are more photons available to interact with the surface, leading to a greater number of photoelectric emissions.
More Electrons Emitted: Each photon that interacts with the photosensitive surface has the potential to eject an electron if its energy exceeds the work function of the material. With higher intensity, more photons possess sufficient energy to overcome the work function and eject electrons from the surface. Therefore, increasing the intensity results in more electrons being emitted, leading to a higher photoelectric current.
Greater Electron Kinetic Energy: As the intensity increases, the number of ejected electrons per unit time also increases. These emitted electrons typically have a range of kinetic energies, depending on the energy of the incident photons. With higher intensity, there is a greater chance of higher-energy photons interacting with the surface, resulting in electrons being emitted with higher kinetic energies. This contributes to an overall increase in the photoelectric current.
Saturated Current: In certain cases, there may be a limit to the increase in photoelectric current with increasing intensity. This is because the photoelectric current can reach a saturation point, where further increases in intensity do not lead to a proportional increase in the number of emitted electrons. At this point, all available electrons in the material have been emitted, and the photoelectric current reaches a maximum value.
Overall, increasing the intensity of monochromatic radiation incident on a photosensitive surface increases the number of photons interacting with the surface, leading to more electrons being ejected and a higher photoelectric current.
Answered on 06 Apr Learn CBSE/Class 12/Science/Physics/Unit 7-Dual Nature of Matter
Sadika
Without access to the given graph, I am unable to directly identify the pairs of curves corresponding to different materials but the same intensity of incident radiation. However, I can provide a general explanation of how you can identify these pairs.
When comparing the photoelectric current versus applied voltage graphs for different photosensitive materials but the same intensity of incident radiation, you need to look for similarities and differences in the characteristics of the curves.
Here's how you can identify the pairs of curves corresponding to different materials but the same intensity of incident radiation:
Similar Characteristics: Look for curves that have similar shapes and trends. Even though the absolute values of the currents may be different, the overall behavior of the curves should be comparable. For example, both curves may exhibit a linear region followed by saturation at higher voltages.
Shifts in Threshold Voltage: Compare the threshold voltages (voltages at which the photoelectric current starts to increase significantly) for each material. If the intensities of the incident radiation are the same, the threshold voltages should be similar for the corresponding pairs of curves. Look for shifts in the threshold voltage between pairs of curves, as these may indicate differences in the work functions of the materials.
Intersections and Crossings: Examine whether the curves intersect or cross each other at any point. If two curves intersect, it suggests that the corresponding materials have different properties. However, if two curves do not intersect but exhibit similar behavior, they likely correspond to different materials but the same intensity of incident radiation.
By carefully analyzing these characteristics, you should be able to identify the pairs of curves that correspond to different materials but the same intensity of incident radiation on the graph.
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