Photoelectric emission is observed from a surface for frequencies `v_1` and `v_2` of the incident radiation (`v_1 gt v_2`) if maximum kinetic energies of the photo electrons in the two cases are in the ratio 1:K, then the threshold frequency is given by :

Options

A. `(v_(2)-v_(1))/(K-1)`
B. `(Kv_(1)-v_(2))/(K-1)`
C. `K(v_(2)-v_(1))/(K-1)`
D. `(v_(2)-v_(1))/(K)`

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, it can eject electrons from that material. The kinetic energy (KE) of the emitted electrons can be expressed as:
\[ KE = h\nu - \phi \]
where:
- \( KE \) is the maximum kinetic energy of the emitted photoelectrons,
- \( h \) is Planck's constant,
- \( \nu \) is the frequency of the incident radiation,
- \( \phi \) is the work function of the material.

### Step 2: Set Up the Equations for Two Frequencies
Given two frequencies \( \nu_1 \) and \( \nu_2 \) (where \( \nu_1 > \nu_2 \)), we can write the kinetic energies for these frequencies as:
\[ KE_1 = h\nu_1 - \phi \]
\[ KE_2 = h\nu_2 - \phi \]

### Step 3: Use the Given Ratio of Kinetic Energies
According to the problem, the maximum kinetic energies are in the ratio \( 1:K \):
\[ \frac{KE_1}{KE_2} = \frac{1}{K} \]

Substituting the expressions for kinetic energy, we have:
\[ \frac{h\nu_1 - \phi}{h\nu_2 - \phi} = \frac{1}{K} \]

### Step 4: Cross Multiply to Eliminate the Fraction
Cross multiplying gives us:
\[ K(h\nu_1 - \phi) = (h\nu_2 - \phi) \]

### Step 5: Rearrange the Equation
Expanding and rearranging the equation:
\[ Kh\nu_1 - K\phi = h\nu_2 - \phi \]
\[ Kh\nu_1 - h\nu_2 = K\phi - \phi \]
\[ Kh\nu_1 - h\nu_2 = \phi(K - 1) \]

### Step 6: Solve for the Work Function \( \phi \)
Now, we can solve for \( \phi \):
\[ \phi = \frac{Kh\nu_1 - h\nu_2}{K - 1} \]

### Step 7: Relate Work Function to Threshold Frequency
The work function \( \phi \) can also be expressed in terms of the threshold frequency \( \nu_0 \):
\[ \phi = h\nu_0 \]
Thus, we can set the two expressions for \( \phi \) equal to each other:
\[ h\nu_0 = \frac{Kh\nu_1 - h\nu_2}{K - 1} \]

### Step 8: Solve for the Threshold Frequency \( \nu_0 \)
Dividing both sides by \( h \):
\[ \nu_0 = \frac{K\nu_1 - \nu_2}{K - 1} \]

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