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

Options

A. `(n_(1)-n_(2))/(k-1)`
B. `(kn_(1)-n_(2))/(k-1)`
C. `(kn_(2)-n_(1))/(k-1)`
D. `(n_(2)-n_(1))/(k)`

Answer

To solve the problem, we will use the photoelectric equation and the given information about the kinetic energy of the emitted photoelectrons.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The kinetic energy (KE) of the emitted photoelectrons can be expressed using the photoelectric equation:
\[ KE = h\nu - \phi \]
where \( KE \) is the kinetic energy of the emitted electrons, \( h \) is Planck's constant, \( \nu \) is the frequency of the incident light, and \( \phi \) is the work function of the metallic surface. The work function can also be expressed in terms of the threshold frequency (\( \nu_t \)):
\[ \phi = h\nu_t \]

2. **Setting Up the Equations**:
For the two frequencies \( \nu_1 \) and \( \nu_2 \) (where \( \nu_1 > \nu_2 \)):
- The maximum kinetic energy for frequency \( \nu_1 \):
\[ KE_1 = h\nu_1 - h\nu_t \]
- The maximum kinetic energy for frequency \( \nu_2 \):
\[ KE_2 = h\nu_2 - h\nu_t \]

3. **Using the Given Ratio of Kinetic Energies**:
We know that the ratio of the maximum kinetic energies is given as:
\[ \frac{KE_1}{KE_2} = \frac{1}{k} \]
Substituting the expressions for \( KE_1 \) and \( KE_2 \):
\[ \frac{h\nu_1 - h\nu_t}{h\nu_2 - h\nu_t} = \frac{1}{k} \]

4. **Simplifying the Equation**:
We can cancel \( h \) from both sides:
\[ \frac{\nu_1 - \nu_t}{\nu_2 - \nu_t} = \frac{1}{k} \]
Cross-multiplying gives:
\[ k(\nu_1 - \nu_t) = \nu_2 - \nu_t \]

5. **Rearranging the Equation**:
Expanding and rearranging the equation:
\[ k\nu_1 - k\nu_t = \nu_2 - \nu_t \]
Bringing all terms involving \( \nu_t \) to one side:
\[ k\nu_1 - \nu_2 = k\nu_t - \nu_t \]
Factoring out \( \nu_t \) on the right side:
\[ k\nu_1 - \nu_2 = \nu_t(k - 1) \]

6. **Solving for the Threshold Frequency**:
Finally, we can solve for the threshold frequency \( \nu_t \):
\[ \nu_t = \frac{k\nu_1 - \nu_2}{k - 1} \]

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