The de Broglie wavelenght `(lambda)` associated with a photoelectron varies with the frequency `(v)` of the incident radiation as,[`v_(0)` is threshold frequency]:

Options

A. `lambda prop (1)/((v-v_(0))^((3)/(2)))`
B. `lambda prop (1)/((v-v_(0))^((1)/(4)))`
C. `prop (1)/((v-v_(0))^((1)/(2)))`
D. `lambda prop (1)/((v-v_(0)))`

Explanation

To solve the problem regarding the de Broglie wavelength (λ) associated with a photoelectron in relation to the frequency (ν) of the incident radiation, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Photoelectric Effect**:
The photoelectric effect states that when light of sufficient frequency (ν) strikes a material, it can eject electrons. The energy of the incident photons is given by the equation:
\[ E = hν \]
where \( h \) is Planck's constant.

2. **Define the Threshold Frequency**:
The threshold frequency (ν₀) is the minimum frequency required to eject an electron from the material. The energy corresponding to this frequency is:
\[ E_0 = hν_0 \]

3. **Energy Conservation**:
The total energy of the incident photon is used to overcome the work function (φ) of the material and to provide kinetic energy (KE) to the ejected electron:
\[ hν = φ + KE \]
where \( KE = \frac{1}{2} mv^2 \) (m is the mass of the electron and v is its velocity).

4. **Express Kinetic Energy**:
Rearranging the energy equation gives:
\[ KE = hν - hν_0 \]
Substituting for KE:
\[ \frac{1}{2} mv^2 = hν - hν_0 \]

5. **Solve for Velocity (v)**:
From the above equation, we can express the velocity (v) of the photoelectron:
\[ v = \sqrt{\frac{2(hν - hν_0)}{m}} \]

6. **Relate Velocity to de Broglie Wavelength**:
The de Broglie wavelength (λ) is given by:
\[ λ = \frac{h}{p} \]
where \( p \) is the momentum. The momentum \( p \) can be expressed as:
\[ p = mv \]
Therefore, substituting for v:
\[ λ = \frac{h}{mv} \]

7. **Substitute for v**:
Substituting the expression for v into the equation for λ:
\[ λ = \frac{h}{m \sqrt{\frac{2(hν - hν_0)}{m}}} \]
Simplifying this gives:
\[ λ = \frac{h \sqrt{m}}{\sqrt{2(hν - hν_0)}} \]
This can be rewritten as:
\[ λ \propto \frac{1}{\sqrt{hν - hν_0}} \]

8. **Final Expression**:
Since \( hν - hν_0 = h(ν - ν_0) \), we can express the relationship as:
\[ λ \propto \frac{1}{\sqrt{ν - ν_0}} \]
Thus, the de Broglie wavelength (λ) is inversely proportional to the square root of the difference between the frequency of the incident radiation and the threshold frequency.

### Conclusion:
The correct relationship is:
\[ λ \propto \frac{1}{\sqrt{ν - ν_0}} \]
Thus, the answer is option C, which indicates that the de Broglie wavelength is inversely proportional to the square root of \( ν - ν_0 \).