Question

Which of the following graphs shows the variation of photoelectric current \(I\) with the intensity of light?

• A.

  

• B.

  

• C.

  

• D.

  

• A. increases linearly
• B. decreases
• C. increases non-linearly
• D. remains the same
• A. Photocurrent versus Intensity of incident light
• B. Photocurrent versus Frequency of incident light
• C. Cut-off potential versus Frequency of incident light
• D. Cut-off potential versus Intensity of incident light
• A. \(K_R>K_Y>K_B\)
• B. \(K_Y>K_B>K_R\)
• C. \(K_B>K_Y>K_R\)
• D. \(K_R>K_B>K_Y\)
• A. Caesium
• B. Zinc
• C. Cadmium
• D. Magnesium

Answer 1

The Correct Option is A

In the photoelectric effect, when light of frequency higher than the threshold frequency is incident on the metal surface, photoelectrons are ejected. The number of photoelectrons emitted is directly proportional to the intensity of light because intensity represents the number of photons striking the surface per unit time. The photoelectric current \(I\) is proportional to the number of photoelectrons emitted, which in turn is directly proportional to the intensity of light. Thus, the graph of photoelectric current \(I\) with intensity of light is a straight line, as shown in Option (A).

Answer 2

In the photoelectric effect, the saturation current depends on the number of photoelectrons emitted, which is determined by the intensity of the light. The saturation current is independent of the frequency of the incident light, as long as the frequency is above the threshold frequency required to eject electrons. When the frequency of the light is increased (above the threshold frequency) without changing its intensity, the energy of each photon increases, but the number of photons (and hence the number of ejected electrons) remains the same because the intensity (which is related to the number of photons) is unchanged. As a result, the saturation current remains the same. Thus, the correct answer is (D): The saturation current remains the same.

Answer 3

According to Einstein’s photoelectric equation: \[ E_k = hf - \phi \] Where: 
- \( E_k \) is the kinetic energy of the emitted electrons, 
- \( h \) is Planck’s constant, 
- \( f \) is the frequency of the incident light, 
- \( \phi \) is the work function of the material. The cut-off potential \( V_0 \) is related to the maximum kinetic energy by: \[ E_k = e V_0 \] Thus, the cut-off potential is proportional to the frequency of the incident light. The graph of \( V_0 \) (cut-off potential) versus \( f \) (frequency) will be a straight line, and the slope of this line will give the value of Planck’s constant \( h \).

Answer 4

According to Einstein's photoelectric equation: \[ K_{\text{max}} = h \nu - \phi \] where \(K_{\text{max}}\) is the maximum kinetic energy of the emitted photoelectrons, \(h\) is Planck's constant, \(\nu\) is the frequency of the incident light, and \(\phi\) is the work function of the metal. Since the intensity of the light is the same for all three colors, the only factor influencing the kinetic energy is the frequency of the incident light. The energy of a photon is given by \(E = h \nu\), so: 
- Red light has the lowest frequency, hence the lowest energy per photon, and thus the lowest kinetic energy of the emitted electrons. 
- Yellow light has a higher frequency and energy per photon compared to red, so the kinetic energy will be higher than for red. 
- Blue light has the highest frequency and energy per photon, so the kinetic energy of the emitted photoelectrons will be the highest. Thus, the kinetic energies follow the order: \[ K_B>K_Y>K_R \] Therefore, the correct answer is (C): \(K_B>K_Y>K_R\).

Answer 5

Among the given metals, **Caesium** (option A) exhibits the photoelectric effect with visible light. The work function of cesium is low enough that visible light has sufficient energy to dislodge electrons from its surface. For the other metals listed (Zinc, Cadmium, and Magnesium), the work function is higher than that of visible light, so they do not exhibit the photoelectric effect with visible light. Therefore, the correct answer is **Caesium**.