Photoelectric emmision is observed from a surface when lights of frequency `n_(1)`and `n_(2)` incident.If the ratio of maximum kinetic energy in two cases is `K: 1` then ( Assume `n_(1) gt n_(2))` threshod frequency is

Options

A. `(K-1)xx(Kn_(2) -n_(1))`
B. `(Kn_(1)-n_(2))/(1-K)`
C. `(K-1)/(Kn_(1)-n_(2))`
D. `(Kn_(2)-n_(1))/(K-1)`

Answer

To solve the question regarding the photoelectric emission and the threshold frequency, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The maximum kinetic energy (K.E.) of the emitted electrons can be expressed as:
\[ K.E. = h\nu - h\nu_0 \]
where \( \nu \) is the frequency of the incident light, \( h \) is Planck's constant, and \( \nu_0 \) is the threshold frequency.

2. **Setting Up the Equations**:
For two frequencies \( \nu_1 \) and \( \nu_2 \):
- For frequency \( \nu_1 \):
\[ K.E_1 = h\nu_1 - h\nu_0 \]
- For frequency \( \nu_2 \):
\[ K.E_2 = h\nu_2 - h\nu_0 \]

3. **Expressing the Ratio of Kinetic Energies**:
Given that the ratio of maximum kinetic energies is \( K:1 \):
\[ \frac{K.E_1}{K.E_2} = K \implies \frac{h\nu_1 - h\nu_0}{h\nu_2 - h\nu_0} = K \]

4. **Simplifying the Equation**:
We can cancel \( h \) from both sides:
\[ \frac{\nu_1 - \nu_0}{\nu_2 - \nu_0} = K \]

5. **Cross Multiplying**:
Cross multiplying gives us:
\[ K(\nu_2 - \nu_0) = \nu_1 - \nu_0 \]

6. **Rearranging the Equation**:
Rearranging the equation leads to:
\[ K\nu_2 - K\nu_0 = \nu_1 - \nu_0 \]
Bringing similar terms together:
\[ K\nu_2 - \nu_1 = K\nu_0 - \nu_0 \]
This simplifies to:
\[ K\nu_2 - \nu_1 = \nu_0(K - 1) \]

7. **Solving for Threshold Frequency**:
Now, we can express the threshold frequency \( \nu_0 \):
\[ \nu_0 = \frac{K\nu_2 - \nu_1}{K - 1} \]

### Final Answer:
The threshold frequency \( \nu_0 \) is given by:
\[ \nu_0 = \frac{K\nu_2 - \nu_1}{K - 1} \]