Photoelectric emission is observed from a metallic surface for frequencies `v_(1) and v_(2)` of the incident light. If the maximum value of kinetic energies of the photoelectrons emitted in the two cases are in the ratio `n:1` then the threshold frequency of the metallic surface is

Options

A. `(v_(2)-v_(1))/(n-1)`
B. `(nv_(2)-v_(1))/(n-1)`
C. `(nv_(1)-v_(2))/(n-1)`
D. `(v_(2)-v_(1))/(n)`

Explanation

To solve the problem of finding the threshold frequency (\( \nu_0 \)) of a metallic surface from the given conditions, we will use the principles of the photoelectric effect and the Einstein's photoelectric equation.

### Step-by-Step Solution:

1. **Understand the Photoelectric Effect**: The maximum kinetic energy (\( KE_{max} \)) of the emitted photoelectrons can be expressed using Einstein's photoelectric equation:
\[ KE_{max} = h\nu - h\nu_0 \]
where \( h \) is Planck's constant, \( \nu \) is the frequency of the incident light, and \( \nu_0 \) is the threshold frequency.

2. **Set Up Equations for Two Frequencies**: For the two frequencies \( \nu_1 \) and \( \nu_2 \), we can write:
- For frequency \( \nu_1 \):
\[ KE_{max,1} = h\nu_1 - h\nu_0 \]
- For frequency \( \nu_2 \):
\[ KE_{max,2} = h\nu_2 - h\nu_0 \]

3. **Express Kinetic Energies**: Let the maximum kinetic energies be represented as:
\[ KE_{max,1} = K_1 \quad \text{and} \quad KE_{max,2} = K_2 \]
Given that the ratio of the maximum kinetic energies is \( n:1 \):
\[ \frac{K_1}{K_2} = n \quad \Rightarrow \quad K_1 = nK_2 \]

4. **Substitute Kinetic Energies into Equations**:
- From the first equation:
\[ K_1 = h\nu_1 - h\nu_0 \quad \Rightarrow \quad nK_2 = h\nu_1 - h\nu_0 \]
- From the second equation:
\[ K_2 = h\nu_2 - h\nu_0 \]

5. **Substitute \( K_2 \) into the First Equation**:
\[ n(h\nu_2 - h\nu_0) = h\nu_1 - h\nu_0 \]

6. **Simplify the Equation**:
\[ nh\nu_2 - nh\nu_0 = h\nu_1 - h\nu_0 \]
Rearranging gives:
\[ nh\nu_2 - h\nu_1 = nh\nu_0 - h\nu_0 \]
\[ nh\nu_2 - h\nu_1 = h(n-1)\nu_0 \]

7. **Solve for Threshold Frequency \( \nu_0 \)**:
\[ \nu_0 = \frac{n\nu_2 - \nu_1}{n - 1} \]

### Final Answer:
The threshold frequency of the metallic surface is:
\[ \nu_0 = \frac{n\nu_2 - \nu_1}{n - 1} \]