When incident wavelength is `lamda`, stopping potential is ` 3V_0`. If incident wavelength is ` 2 lamda ` then stopping potential is ` V_0`. Find out threshold wavelength in terms of ` lamda`.

Answer

To solve the problem step by step, we will use Einstein's photoelectric equation and the information given in the question.

### Step 1: Understand the Problem
We are given two scenarios:
1. When the incident wavelength is \( \lambda \), the stopping potential is \( 3V_0 \).
2. When the incident wavelength is \( 2\lambda \), the stopping potential is \( V_0 \).

We need to find the threshold wavelength \( \lambda_0 \) in terms of \( \lambda \).

### Step 2: Write Einstein's Photoelectric Equation
Einstein's photoelectric equation relates the stopping potential \( V \), the energy of the incident photon, and the work function \( \phi \) of the material:
\[ eV = \frac{hc}{\lambda} - \phi \]
where:
- \( e \) is the charge of an electron,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the incident light,
- \( \phi \) is the work function of the material.

### Step 3: Set Up Equations for Both Cases
**For the first case (wavelength \( \lambda \), stopping potential \( 3V_0 \)):**
\[ 3eV_0 = \frac{hc}{\lambda} - \phi \quad \text{(1)} \]

**For the second case (wavelength \( 2\lambda \), stopping potential \( V_0 \)):**
\[ eV_0 = \frac{hc}{2\lambda} - \phi \quad \text{(2)} \]

### Step 4: Rearranging the Equations
From equation (1):
\[ \phi = \frac{hc}{\lambda} - 3eV_0 \quad \text{(3)} \]

From equation (2):
\[ \phi = \frac{hc}{2\lambda} - eV_0 \quad \text{(4)} \]

### Step 5: Equate the Two Expressions for Work Function
Set equations (3) and (4) equal to each other:
\[ \frac{hc}{\lambda} - 3eV_0 = \frac{hc}{2\lambda} - eV_0 \]

### Step 6: Solve for \( eV_0 \)
Rearranging gives:
\[ \frac{hc}{\lambda} - \frac{hc}{2\lambda} = 3eV_0 - eV_0 \]
\[ \frac{hc}{\lambda} - \frac{hc}{2\lambda} = 2eV_0 \]
Finding a common denominator:
\[ \frac{2hc - hc}{2\lambda} = 2eV_0 \]
\[ \frac{hc}{2\lambda} = 2eV_0 \]
\[ eV_0 = \frac{hc}{4\lambda} \quad \text{(5)} \]

### Step 7: Substitute \( eV_0 \) Back into One of the Equations
Using equation (1):
\[ 3eV_0 = \frac{hc}{\lambda} - \phi \]
Substituting \( eV_0 \) from equation (5):
\[ 3 \left(\frac{hc}{4\lambda}\right) = \frac{hc}{\lambda} - \phi \]
\[ \frac{9hc}{4\lambda} = \frac{hc}{\lambda} - \phi \]
Rearranging gives:
\[ \phi = \frac{hc}{\lambda} - \frac{9hc}{4\lambda} \]
Finding a common denominator:
\[ \phi = \frac{4hc - 9hc}{4\lambda} = \frac{-5hc}{4\lambda} \]

### Step 8: Find the Threshold Wavelength
Using the expression for work function:
\[ \phi = \frac{hc}{\lambda_0} \]
Setting the two expressions for \( \phi \) equal gives:
\[ \frac{hc}{\lambda_0} = \frac{5hc}{4\lambda} \]
Cancelling \( hc \) from both sides:
\[ \frac{1}{\lambda_0} = \frac{5}{4\lambda} \]
Taking the reciprocal:
\[ \lambda_0 = \frac{4\lambda}{5} \]

### Final Answer
The threshold wavelength \( \lambda_0 \) in terms of \( \lambda \) is:
\[ \lambda_0 = \frac{4\lambda}{5} \]