Question

The frequency and the intensity of a beam of light falling on the surface of photoelectric material are increased by a factor of two. Treating efficiency of photoelectron generation as constant, this will:

• A. increase the maximum energy of the photoelectrons, as well as photoelectric current by a factor of two.
• B. increase the maximum kinetic energy of the photo electrons and would increase the photoelectric current by a factor of two.
• C. increase the maximum kinetic energy of the photoelectrons by a factor of greater than two and will have no effect on the magnitude of photoelectric current produced.
• D. not produce any effect on the kinetic energy of the emitted electrons but will increase the photoelectric current by a factor of two.

Answer 1

Answer: (B)

increase the maximum kinetic energy of the photo electrons and would increase the photoelectric current by a factor of two.

Answer 2

The photoelectric effect describes the emission of electrons when light shines on a material. Let's analyze the effect of increasing frequency and intensity separately.

1. Effect of Frequency on Maximum Kinetic Energy ($K_{max}$): According to Einstein's photoelectric equation: $K_{max} = hν - Φ$ where $h$ is Planck's constant, $ν$ is the frequency of incident light, and $Φ$ is the work function of the material.

   Initially, let the frequency be $ν_1$ and the maximum kinetic energy be $K_{max,1}$: $K_{max,1} = hν_1 - Φ$

   When the frequency is increased by a factor of two, the new frequency is $ν_2 = 2ν_1$. The new maximum kinetic energy will be $K_{max,2}$: $K_{max,2} = h(2ν_1) - Φ = 2hν_1 - Φ$

   We can rewrite $K_{max,2}$ in terms of $K_{max,1}$: $K_{max,2} = 2hν_1 - Φ = 2(hν_1 - Φ) + 2Φ - Φ = 2K_{max,1} + Φ$

   Since the work function $Φ$ is always positive for a material ($Φ > 0$), it follows that: $K_{max,2} > 2K_{max,1}$

   This means the maximum kinetic energy of the photoelectrons increases by a factor greater than two.

2. Effect of Intensity on Photoelectric Current: The intensity of light is proportional to the number of photons incident per unit area per unit time. In the photoelectric effect, each photon (with energy greater than the work function) can eject one electron.

   The photoelectric current is directly proportional to the number of photoelectrons emitted per unit time.

   The problem states that the intensity of the beam of light is increased by a factor of two, and the efficiency of photoelectron generation is constant. This means that if the number of incident photons doubles, the number of emitted photoelectrons also doubles.

   Therefore, the photoelectric current will increase by a factor of two.

Conclusion: The maximum kinetic energy increases (specifically, by a factor greater than two), and the photoelectric current increases by a factor of two. Option (B) accurately reflects both these changes.