Answer:

Introduction:
In a photo cell, when light of a specific wavelength (lambda) falls on the surface, it excites electrons and causes them to move. The speed of the fastest electron depends on the wavelength of the incident light. In this case, we are given that the excitation wavelength is changed to 3lambda/4, and we need to determine the speed of the fastest electron in this scenario.

Explanation:
To understand the relationship between the wavelength and the speed of the fastest electron, let's consider the photoelectric effect and the energy of a photon.

- The photoelectric effect states that when light falls on a metal surface, electrons are emitted if the light has sufficient energy.
- The energy of a photon can be calculated using the equation E = hc/lambda, where E is the energy, h is Planck's constant, c is the speed of light, and lambda is the wavelength of the incident light.
- The kinetic energy of the emitted electron can be calculated using the equation KE = 1/2 mv^2, where KE is the kinetic energy, m is the mass of the electron, and v is its velocity.

Effect of changing the wavelength:
When the excitation wavelength is changed to 3lambda/4, the energy of the incident photons will be different. Let's compare the energy of the incident photons in the two scenarios.

1. When the excitation wavelength is lambda:
- Energy of the incident photons: E1 = hc/lambda
- Kinetic energy of the fastest electron: KE1 = 1/2 m v^2

2. When the excitation wavelength is 3lambda/4:
- Energy of the incident photons: E2 = hc/(3lambda/4) = 4hc/3lambda
- Kinetic energy of the fastest electron: KE2 = 1/2 m v2^2

Comparing the speeds:
To compare the speeds of the fastest electron in the two scenarios, we can equate the kinetic energies:

KE1 = KE2
1/2 m v^2 = 1/2 m v2^2

Cancelling the mass (m) from both sides and rearranging the equation, we get:

v^2 = v2^2

Taking the square root of both sides, we get:

v = v2

Conclusion:
From the above comparison, we can conclude that when the excitation wavelength is changed to 3lambda/4, the speed of the fastest electron remains the same as before. Therefore, the answer is option A) v(3/4)^(1/2).

The answer should be D.