In photoelectric effect light of certain frequency `( gt ` threshold frequency ) is incident on a metal surfave whereby , an `e^(-)` ( with certain K.E.) moves towards the collector plate and a flow of current is initiated . In order to stop the current flow, an opposite potential , on the two metal plates, is applied
The work function of a metal is 4eV. If light of frequency `2.3 xx 10^(15) Hz` is incident on metal surface, then

Options

A. No photoelectron will be ejected
B. 2 photoelectron of zero kinetic energy are ejected
C. 1 photoelectron of zero kinetic energy is ejected
D. 1 photoelectron is ejected , which required the stopping potential of 5.52 volt

Answer

To solve the problem related to the photoelectric effect, we will follow these steps:

### Step 1: Understand the given data
- Work function of the metal (Φ) = 4 eV
- Frequency of the incident light (ν) = 2.3 × 10¹⁵ Hz
- Planck's constant (h) = 6.626 × 10⁻³⁴ J·s
- 1 eV = 1.6 × 10⁻¹⁹ J

### Step 2: Calculate the energy of the incident light
The energy of the incident light can be calculated using the formula:
\[ E = hν \]

Substituting the values:
\[ E = 6.626 × 10^{-34} \, \text{J·s} \times 2.3 × 10^{15} \, \text{Hz} \]

Calculating this gives:
\[ E = 1.52 × 10^{-18} \, \text{J} \]

### Step 3: Convert the energy from joules to electron volts
To convert the energy from joules to electron volts, we use the conversion factor:
\[ E \, (\text{in eV}) = \frac{E \, (\text{in J})}{1.6 × 10^{-19} \, \text{J/eV}} \]

Calculating this:
\[ E \, (\text{in eV}) = \frac{1.52 × 10^{-18}}{1.6 × 10^{-19}} \approx 9.5 \, \text{eV} \]

### Step 4: Apply the photoelectric equation
The photoelectric equation is given by:
\[ E = Φ + KE \]
Where:
- \( KE \) is the kinetic energy of the emitted electron.

Rearranging gives:
\[ KE = E - Φ \]

Substituting the values:
\[ KE = 9.5 \, \text{eV} - 4 \, \text{eV} = 5.5 \, \text{eV} \]

### Step 5: Calculate the stopping potential
The stopping potential (Vs) can be calculated using the relationship:
\[ KE = eV_s \]

Where:
- \( e \) is the charge of an electron (1 eV corresponds to 1 V for a single electron).

Thus:
\[ V_s = KE \]

Substituting the value:
\[ V_s = 5.5 \, \text{V} \]

### Conclusion
From the calculations, we find that:
- One photoelectron is ejected which requires a stopping potential of approximately 5.5 V.

### Final Answer
The correct statement is: **1 photoelectron is ejected which requires the stopping potential of 5.52 V.**

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