STATEMENT.1 Photo-omission from a photosensitive surface is possible only if the  incident radiation has a frequency above threshold frequency. STATEMENT-2: Unless, the energy of incident photo is greater than the work function of photo- sensitive surface, no photoemission is possible.

Options

A. STATEMENT-1 is true, STATEMENT -2  is true , STATEMENT 2 is a correct explanation for STATEMENT -1.
B. STATEMENT-1 is true, STATEMENT -2  is true , STATEMENT 2 is NOT a correct explanation for STATEMENT -1.
C. STATEMENT-1 is true, STATEMENT -2  is False.
D. STATEMENT-1 is False, STATEMENT -2  is True

Answer

To analyze the statements regarding photoemission from a photosensitive surface, we can break down the concepts involved and provide a step-by-step solution.

### Step-by-Step Solution:

**Step 1: Understanding Photoemission**
- Photoemission refers to the process where electrons are ejected from a material (typically a metal) when it is exposed to light (or other electromagnetic radiation).

**Step 2: Defining Threshold Frequency**
- The threshold frequency (denoted as \( \nu_0 \)) is the minimum frequency of incident radiation required to eject electrons from the surface of the material. This frequency corresponds to the minimum energy needed to overcome the work function of the material.

**Step 3: Relationship Between Energy and Frequency**
- The energy of a photon is given by the equation:
\[ E = h\nu \]
where \( E \) is the energy of the photon, \( h \) is Planck's constant, and \( \nu \) is the frequency of the incident radiation.

**Step 4: Work Function**
- The work function (\( \phi \)) is the minimum energy required to remove an electron from the surface of the material. It is a characteristic property of the material.

**Step 5: Condition for Photoemission**
- For photoemission to occur, the energy of the incident photon must be greater than or equal to the work function:
\[ h\nu \geq \phi \]
This implies that the frequency of the incident radiation must be at least equal to the threshold frequency:
\[ \nu \geq \nu_0 \]
where \( \nu_0 = \frac{\phi}{h} \).

**Step 6: Evaluating the Statements**
- **Statement 1**: "Photoemission from a photosensitive surface is possible only if the incident radiation has a frequency above threshold frequency."
- This statement is true because if the frequency is below the threshold frequency, the energy of the photons will not be sufficient to overcome the work function.

- **Statement 2**: "Unless the energy of the incident photon is greater than the work function of the photosensitive surface, no photoemission is possible."
- This statement is also true as it directly relates to the condition for photoemission.

**Step 7: Conclusion**
- Both statements are true, and Statement 2 provides the correct explanation for Statement 1. Therefore, the correct option is that both statements are true, and Statement 2 is the reason for Statement 1.