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 problem, we will follow these steps:

### Step 1: Understand the Photoelectric Effect
The photoelectric effect states that when light of a certain frequency strikes a material, electrons are emitted from that material. The energy of the incident light is given by \( E = h\nu \), where \( h \) is Planck's constant and \( \nu \) is the frequency of the light.

### Step 2: Write the Energy Equation
For the photoelectric effect, the energy of the incident light is equal to the maximum kinetic energy of the emitted electrons plus the work function (\( \phi \)) of the material:
\[ E = K.E + \phi \]
This can be rewritten as:
\[ h\nu = K + h\nu_0 \]
where \( \nu_0 \) is the threshold frequency (the minimum frequency needed to emit electrons).

### Step 3: Set Up Equations for Two Frequencies
For the first frequency \( \nu_1 \):
\[ h\nu_1 = K_1 + h\nu_0 \implies K_1 = h\nu_1 - h\nu_0 \]
For the second frequency \( \nu_2 \):
\[ h\nu_2 = K_2 + h\nu_0 \implies K_2 = h\nu_2 - h\nu_0 \]

### Step 4: Express the Kinetic Energies
From the above equations, we can express the maximum kinetic energies:
\[ K_1 = h(\nu_1 - \nu_0) \quad \text{(1)} \]
\[ K_2 = h(\nu_2 - \nu_0) \quad \text{(2)} \]

### Step 5: Set Up the Ratio of Kinetic Energies
We are given that the ratio of maximum kinetic energies is \( K:1 \):
\[ \frac{K_1}{K_2} = K \implies \frac{h(\nu_1 - \nu_0)}{h(\nu_2 - \nu_0)} = K \]
The \( h \) cancels out:
\[ \frac{\nu_1 - \nu_0}{\nu_2 - \nu_0} = K \]

### Step 6: Cross Multiply to Solve for \( \nu_0 \)
Cross-multiplying gives:
\[ \nu_1 - \nu_0 = K(\nu_2 - \nu_0) \]
Expanding this:
\[ \nu_1 - \nu_0 = K\nu_2 - K\nu_0 \]
Rearranging terms:
\[ \nu_1 + (K - 1)\nu_0 = K\nu_2 \]
Now isolate \( \nu_0 \):
\[ (K - 1)\nu_0 = K\nu_2 - \nu_1 \]
\[ \nu_0 = \frac{K\nu_2 - \nu_1}{K - 1} \]

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