When a certain photosensistive surface is illuminated with monochromatic light of frequency v, the stopping potential for the photo current is `V_(0)//2`. When the surface is illuminated by monochromatic light of frequency `v//2`, the stopping potential is `V_(0)`. The threshold frequency gor photoelectric emission is :

Options

A. `(5v)/(3)`
B. `(4)/(3)v`
C. `2v`
D. `(3v)/(2)`

Explanation

To solve the problem, we will use the photoelectric effect equation given by Einstein:

\[ E = h\nu - \Phi \]

where:
- \( E \) is the energy of the emitted photoelectrons,
- \( h \) is Planck's constant,
- \( \nu \) is the frequency of the incident light,
- \( \Phi \) is the work function of the material.

### Step 1: Set up the equations for both cases

1. **For the first case** (frequency \( \nu \) and stopping potential \( V_0/2 \)):
\[ e\left(\frac{V_0}{2}\right) = h\nu - \Phi \]
This can be rewritten as:
\[ eV_0 = 2h\nu - 2\Phi \quad \text{(Equation 1)} \]

2. **For the second case** (frequency \( \nu/2 \) and stopping potential \( V_0 \)):
\[ eV_0 = h\left(\frac{\nu}{2}\right) - \Phi \]
This can be rewritten as:
\[ eV_0 = \frac{h\nu}{2} - \Phi \quad \text{(Equation 2)} \]

### Step 2: Subtract the two equations

Now, we will subtract Equation 2 from Equation 1:

\[ (2h\nu - 2\Phi) - \left(\frac{h\nu}{2} - \Phi\right) = eV_0 - eV_0 \]

This simplifies to:

\[ 2h\nu - 2\Phi - \frac{h\nu}{2} + \Phi = 0 \]

Rearranging gives:

\[ 2h\nu - \frac{h\nu}{2} - \Phi = 0 \]

### Step 3: Combine like terms

Combining the terms involving \( h\nu \):

\[ \left(2 - \frac{1}{2}\right)h\nu - \Phi = 0 \]

This simplifies to:

\[ \frac{4h\nu}{2} - \Phi = 0 \]

### Step 4: Solve for the work function

Rearranging gives:

\[ \Phi = 4h\nu - 2\Phi \]

So,

\[ 3\Phi = 4h\nu \]

### Step 5: Express the work function in terms of threshold frequency

The work function \( \Phi \) can also be expressed in terms of the threshold frequency \( \nu_0 \):

\[ \Phi = h\nu_0 \]

Substituting this into the equation gives:

\[ 3h\nu_0 = 4h\nu \]

### Step 6: Solve for the threshold frequency

Dividing both sides by \( h \):

\[ 3\nu_0 = 4\nu \]

Thus, we can solve for \( \nu_0 \):

\[ \nu_0 = \frac{4\nu}{3} \]

### Final Answer

The threshold frequency for photoelectric emission is:

\[ \nu_0 = \frac{4\nu}{3} \]