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 metal is 6eV . If light of frequency ` 1 xx 10^(15) `Hz is incident on the metal , intensity of light is increased 4 times, then

Options

A. No photoelectron will be ejected
B. 8 photoelectrons of zero kinetic energy shall be ejected
C. 2 photoelectrons of 2 eV kinetic energy are ejected
D. Only one photoelectron is ejected

Answer

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

### Step 1: Understand the given data
- Work function (φ) of the metal = 6 eV
- Frequency of the incident light (ν) = 1 x 10^15 Hz
- Intensity of light is increased 4 times.

### Step 2: Convert the work function from eV to Joules
The work function in Joules can be calculated using the conversion factor:
1 eV = 1.6 x 10^-19 Joules.

So,
\[ \phi = 6 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 9.6 \times 10^{-19} \, \text{J} \]

### Step 3: Calculate the threshold frequency (ν₀)
The work function is related to the threshold frequency by the equation:
\[ \phi = h \nu_0 \]
where \(h\) (Planck's constant) = 6.63 x 10^-34 J·s.

Rearranging the equation gives:
\[ \nu_0 = \frac{\phi}{h} = \frac{9.6 \times 10^{-19} \, \text{J}}{6.63 \times 10^{-34} \, \text{J·s}} \]

Calculating this gives:
\[ \nu_0 \approx 1.44 \times 10^{15} \, \text{Hz} \]

### Step 4: Compare the incident frequency with the threshold frequency
- Incident frequency (ν) = 1 x 10^15 Hz
- Threshold frequency (ν₀) = 1.44 x 10^15 Hz

Since the incident frequency (1 x 10^15 Hz) is less than the threshold frequency (1.44 x 10^15 Hz), no electrons will be emitted from the metal surface.

### Step 5: Conclusion
Even if the intensity of light is increased by 4 times, it does not affect the emission of electrons because the frequency of the incident light is still below the threshold frequency. Therefore, the correct answer is:

**No photoelectrons will be ejected.**