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

When light with a frequency exceeding the threshold frequency strikes a metal surface, the photoelectric effect causes photoelectrons to be ejected. The quantity of emitted photoelectrons is directly proportional to the light's intensity, as intensity quantifies the number of photons impacting the surface per unit time. The photoelectric current \(I\) is directly proportional to the number of photoelectrons emitted, which, consequently, is directly proportional to the light's intensity. Therefore, a plot of photoelectric current \(I\) against light intensity yields a straight line, consistent with Option (A).

Answer 2

In the photoelectric effect, saturation current is proportional to the number of emitted photoelectrons, which is directly dictated by light intensity. Crucially, this current is unaffected by incident light frequency, provided it exceeds the minimum frequency required for electron ejection. Increasing the light frequency (above the threshold) while keeping intensity constant elevates individual photon energy but does not alter the photon count, and thus the ejected electron count, due to the constant intensity. Consequently, the saturation current remains constant. Therefore, the correct conclusion is that the saturation current remains unchanged.

Answer 3

Einstein's photoelectric equation states: \[ E_k = hf - \phi \], where \( E_k \) represents the kinetic energy of the emitted electrons, \( h \) is Planck’s constant, \( f \) is the frequency of the incident light, and \( \phi \) is the work function of the material. The cut-off potential \( V_0 \) is linked to the maximum kinetic energy by the relation: \[ E_k = e V_0 \]. Consequently, the cut-off potential is directly proportional to the incident light's frequency. A plot of \( V_0 \) (cut-off potential) against \( f \) (frequency) will yield a straight line, with the slope of this line equaling Planck’s constant \( h \).

Answer 4

Einstein's photoelectric equation states: \[ K_{\text{max}} = h u - \phi \]. Here, \(K_{\text{max}}\) represents the maximum kinetic energy of photoelectrons, \(h\) is Planck's constant, \(u\) is the incident light frequency, and \(\phi\) is the metal's work function. Given constant light intensity across all colors, only incident light frequency affects kinetic energy. Photon energy is \(E = h u\). Consequently:
- Red light exhibits the lowest frequency and thus the lowest photon energy, resulting in the lowest kinetic energy for emitted electrons. 
- Yellow light, with its higher frequency and photon energy than red, will yield higher kinetic energy. 
- Blue light possesses the highest frequency and photon energy, leading to the maximum kinetic energy for photoelectrons. Therefore, the kinetic energies are ordered as: \[ K_B>K_Y>K_R \] This establishes option (C) as the correct answer: \(K_B>K_Y>K_R\).

Answer 5

Caesium (option A) is the metal among those provided that exhibits the photoelectric effect with visible light. Its work function is sufficiently low, allowing visible light to possess enough energy to eject electrons from its surface. Conversely, Zinc, Cadmium, and Magnesium have work functions higher than the energy of visible light, thus not exhibiting the photoelectric effect under these conditions. Consequently, Caesium is identified as the correct answer.