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

Explanation

To solve the problem, we will use the principles of the photoelectric effect. The key points to remember are:

1. The energy of the incident photon is given by \( E = h \nu \), where \( h \) is Planck's constant and \( \nu \) is the frequency of the incident radiation.
2. The maximum kinetic energy (KE) of the emitted photoelectrons can be expressed as:
\[ KE = h \nu - \phi \]
where \( \phi \) is the work function of the material, which is related to the threshold frequency \( \nu_0 \) by \( \phi = h \nu_0 \).

### Step-by-step Solution:

1. **Write the equations for the maximum kinetic energies:**
For the two frequencies \( \nu_1 \) and \( \nu_2 \):
\[ KE_1 = h \nu_1 - h \nu_0 \quad (1) \]
\[ KE_2 = h \nu_2 - h \nu_0 \quad (2) \]

2. **Express the kinetic energies in terms of the threshold frequency:**
From equations (1) and (2), we can express the kinetic energies as:
\[ KE_1 = h (\nu_1 - \nu_0) \]
\[ KE_2 = h (\nu_2 - \nu_0) \]

3. **Set up the ratio of the 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 \( KE_2 \):
\[ \frac{h (\nu_1 - \nu_0)}{h (\nu_2 - \nu_0)} = \frac{1}{K} \]

4. **Simplify the equation:**
The \( h \) cancels out:
\[ \frac{\nu_1 - \nu_0}{\nu_2 - \nu_0} = \frac{1}{K} \]
Cross-multiplying gives:
\[ K (\nu_1 - \nu_0) = \nu_2 - \nu_0 \]

5. **Rearranging for the threshold frequency \( \nu_0 \):**
Rearranging the equation:
\[ K \nu_1 - K \nu_0 = \nu_2 - \nu_0 \]
\[ K \nu_1 - \nu_2 = K \nu_0 - \nu_0 \]
\[ K \nu_1 - \nu_2 = (K - 1) \nu_0 \]
Finally, solving for \( \nu_0 \):
\[ \nu_0 = \frac{K \nu_1 - \nu_2}{K - 1} \]

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