Electrons ejected from the surface of a metal, when light of certain frequency is incident on it, are stopped fully by a retarding potential of 3 volts. Photo electric effect in this metallic surface begains at a frequency `6xx10^(14)s^(-1)`. The frequency of the incident light in `s^(-1)` is `[h=6 x10^(-34) J-sec`, charge on the electron =`1.6xx10^(-19)C`]

Options

A. `7.5xx10^(13)`
B. `13.5xx10^(13)`
C. `14xx10^(14)`
D. `7.5xx10^(15)`

Explanation

To solve the problem, we need to determine the frequency of the incident light that causes the photoelectric effect in the metal. We will use the photoelectric effect equation and the information given in the question.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The photoelectric effect states that when light of a certain frequency shines on a metal surface, electrons are ejected. The kinetic energy (KE) of the ejected electrons can be expressed as:
\[ KE = hf - \phi \]
where \( h \) is Planck's constant, \( f \) is the frequency of the incident light, and \( \phi \) is the work function of the metal.

2. **Relating Stopping Potential to Kinetic Energy**:
The stopping potential \( V \) is related to the maximum kinetic energy of the ejected electrons. The kinetic energy can also be expressed in terms of the stopping potential:
\[ KE = eV \]
where \( e \) is the charge of the electron and \( V \) is the stopping potential.

3. **Calculating the Kinetic Energy**:
Given that the stopping potential \( V = 3 \) volts and the charge of the electron \( e = 1.6 \times 10^{-19} \) C, we can calculate the kinetic energy:
\[ KE = eV = (1.6 \times 10^{-19} \, \text{C})(3 \, \text{V}) = 4.8 \times 10^{-19} \, \text{J} \]

4. **Finding the Work Function**:
The photoelectric effect begins at a frequency \( f_0 = 6 \times 10^{14} \, \text{s}^{-1} \). The work function \( \phi \) can be calculated using:
\[ \phi = hf_0 \]
where \( h = 6 \times 10^{-34} \, \text{J-sec} \). Thus,
\[ \phi = (6 \times 10^{-34} \, \text{J-sec})(6 \times 10^{14} \, \text{s}^{-1}) = 3.6 \times 10^{-19} \, \text{J} \]

5. **Finding the Frequency of Incident Light**:
Now, substituting the values of \( KE \) and \( \phi \) into the photoelectric equation:
\[ KE = hf - \phi \]
We can rearrange this to find \( f \):
\[ f = \frac{KE + \phi}{h} \]
Substituting the values we have:
\[ f = \frac{(4.8 \times 10^{-19} \, \text{J}) + (3.6 \times 10^{-19} \, \text{J})}{6 \times 10^{-34} \, \text{J-sec}} = \frac{8.4 \times 10^{-19} \, \text{J}}{6 \times 10^{-34} \, \text{J-sec}} = 1.4 \times 10^{15} \, \text{s}^{-1} \]

### Final Answer:
The frequency of the incident light is \( 1.4 \times 10^{15} \, \text{s}^{-1} \).