In photoelectric effect when photons of energy hv fall on a photosensitive surface (work function hv^) electrons are emitted from the metallic surface with a kinetic energy. It is possible to say that:

Options

A. All ejected electrons have same kinetic energy equal to ` hv-hv_(0)`
B. The ejected electrons have a distribution of kinetic energy from zero to ` (hv - hv_(0))`
C. The most energetic electrons have kinetic energy equal to hv
D. All ejected electrons have kinetic energy ` hv_(0)`

Answer

To solve the question regarding the photoelectric effect, we will follow these steps:

### Step 1: Understand the Photoelectric Effect
The photoelectric effect occurs when light (photons) strikes a photosensitive surface (like a metal), causing the emission of electrons. The energy of the incident photons is given by \( E = h\nu \), where \( h \) is Planck's constant and \( \nu \) is the frequency of the light.

### Step 2: Define Work Function
The work function (\( \phi \)) is the minimum energy required to remove an electron from the surface of the metal. It is denoted as \( \phi = h\nu_0 \), where \( \nu_0 \) is the threshold frequency.

### Step 3: Apply the Photoelectric Equation
The relationship between the energy of the incoming photons, the work function, and the kinetic energy of the emitted electrons is given by the equation:
\[ h\nu = h\nu_0 + KE \]
where \( KE \) is the kinetic energy of the emitted electrons.

### Step 4: Rearranging the Equation
From the equation, we can express the kinetic energy of the emitted electrons as:
\[ KE = h\nu - h\nu_0 \]
This shows that the kinetic energy of the emitted electrons depends on the difference between the energy of the incoming photons and the work function.

### Step 5: Analyze the Kinetic Energy Range
The kinetic energy of the emitted electrons can vary from \( 0 \) (when \( h\nu = h\nu_0 \)) to a maximum value of \( h\nu - h\nu_0 \) (when the energy of the incoming photon is greater than the work function). Thus, the kinetic energy of the emitted electrons can be in the range:
\[ 0 \leq KE \leq h\nu - h\nu_0 \]

### Conclusion
Based on the analysis, we can conclude that when photons of energy \( h\nu \) fall on a photosensitive surface with work function \( h\nu_0 \), the emitted electrons will have a distribution of kinetic energy ranging from \( 0 \) to \( h\nu - h\nu_0 \).

### Final Answer
The correct statement regarding the emitted electrons is that they have a distribution of kinetic energy from \( 0 \) to \( h\nu - h\nu_0 \).

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