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
A light of frequency `2.5xx 10^(15) Hz` is incident on a metal surface having work function 4 eV. The velocity of photoelectron is ( in cm `s^(-1))`

Options

A. `1.5 xx 10^(6)`
B. `1.5 xx 10^(8)`
C. `2 xx 10^(8)`
D. `2.5 xx 10^(4)`

Answer

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

### Step 1: Understand the given data
- Frequency of incident light, \( \nu = 2.5 \times 10^{15} \, \text{Hz} \)
- Work function of the metal, \( \phi = 4 \, \text{eV} \)

### Step 2: Convert the work function from eV to Joules
The work function in Joules can be calculated using the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ \phi = 4 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 6.4 \times 10^{-19} \, \text{J} \]

### Step 3: Calculate the energy of the incident photons
The energy of the incident photons can be calculated using the formula:
\[ E = h \nu \]
Where \( h = 6.626 \times 10^{-34} \, \text{J s} \) (Planck's constant). Thus,
\[ E = 6.626 \times 10^{-34} \, \text{J s} \times 2.5 \times 10^{15} \, \text{Hz} = 1.6565 \times 10^{-18} \, \text{J} \]

### Step 4: Calculate the kinetic energy of the emitted photoelectron
The kinetic energy (K.E.) of the emitted photoelectron can be calculated using the equation:
\[ \text{K.E.} = E - \phi \]
Substituting the values:
\[ \text{K.E.} = 1.6565 \times 10^{-18} \, \text{J} - 6.4 \times 10^{-19} \, \text{J} = 1.0165 \times 10^{-18} \, \text{J} \]

### Step 5: Relate kinetic energy to velocity
The kinetic energy of the emitted electron is also given by:
\[ \text{K.E.} = \frac{1}{2} mv^2 \]
Where \( m \) is the mass of the electron, \( m = 9.1 \times 10^{-31} \, \text{kg} \). Rearranging for \( v \):
\[ v = \sqrt{\frac{2 \times \text{K.E.}}{m}} \]
Substituting the values:
\[ v = \sqrt{\frac{2 \times 1.0165 \times 10^{-18} \, \text{J}}{9.1 \times 10^{-31} \, \text{kg}}} \]

### Step 6: Calculate the velocity
Calculating the above expression:
\[ v = \sqrt{\frac{2.033 \times 10^{-18}}{9.1 \times 10^{-31}}} = \sqrt{2.23 \times 10^{12}} \approx 1.49 \times 10^6 \, \text{m/s} \]

### Step 7: Convert the velocity to cm/s
To convert from meters per second to centimeters per second:
\[ v = 1.49 \times 10^6 \, \text{m/s} \times 100 = 1.49 \times 10^8 \, \text{cm/s} \]

### Final Answer
The velocity of the photoelectron is approximately \( 1.49 \times 10^8 \, \text{cm/s} \).

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