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. `(v_(1)-v_(2))/(k-1)`
B. `(kv_(1)-v_(2))/(k-1)`
C. `(kv_(2)-v_(1))/(k-1)`
D. `(v_(2)-v_(1))/(k)`

Answer

To solve the problem of determining the threshold frequency of a metallic surface based on the given conditions of photoelectric emission, we can follow these steps:

### Step 1: Understand the Photoelectric Effect
The photoelectric effect states that when light of a certain frequency shines on a metallic surface, it can emit electrons. The kinetic energy (KE) of the emitted electrons can be expressed using Einstein's photoelectric equation:
\[ KE = h\nu - W_0 \]
where:
- \( KE \) is the maximum kinetic energy of the emitted photoelectrons,
- \( h \) is Planck's constant,
- \( \nu \) is the frequency of the incident light,
- \( W_0 \) is the work function of the metallic surface.

### Step 2: Write the Equations for Two Frequencies
For the two frequencies \( \nu_1 \) and \( \nu_2 \) (where \( \nu_1 > \nu_2 \)), we can write the equations for the maximum kinetic energies:
1. For frequency \( \nu_1 \):
\[ KE_1 = h\nu_1 - W_0 \]
2. For frequency \( \nu_2 \):
\[ KE_2 = h\nu_2 - W_0 \]

### Step 3: Set Up the 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 \( KE_1 \) and \( KE_2 \):
\[ \frac{h\nu_1 - W_0}{h\nu_2 - W_0} = \frac{1}{k} \]

### Step 4: Cross-Multiply to Eliminate the Fraction
Cross-multiplying gives:
\[ k(h\nu_1 - W_0) = h\nu_2 - W_0 \]

### Step 5: Rearrange the Equation
Rearranging the equation to isolate \( W_0 \):
\[ kh\nu_1 - kW_0 = h\nu_2 - W_0 \]
\[ kh\nu_1 - h\nu_2 = kW_0 - W_0 \]
Factoring out \( W_0 \) on the right side:
\[ kh\nu_1 - h\nu_2 = W_0(k - 1) \]

### Step 6: Solve for the Work Function
Now, we can express the work function \( W_0 \):
\[ W_0 = \frac{h\nu_1 - h\nu_2}{k - 1} \]

### Step 7: Relate Work Function to Threshold Frequency
The work function \( W_0 \) can also be expressed in terms of the threshold frequency \( \nu_0 \):
\[ W_0 = h\nu_0 \]
Thus, we can equate:
\[ h\nu_0 = \frac{h\nu_1 - h\nu_2}{k - 1} \]

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

### Final Result
Thus, the threshold frequency \( \nu_0 \) of the metallic surface is:
\[ \nu_0 = \frac{\nu_1 - \nu_2}{k - 1} \]