If a certain metal was irradiated by using two different light radiations of frequency 'x' and '2x' the maximum kinetic energy of the ejected electrons are 'y' and '3y' respectively the threshold frequency of the metal is

Options

A. `(x)/(3)`
B. `(x)/(2)`
C. `(3x)/(2)`
D. `(2x)/(3)`

Answer

To solve the problem, we will use the photoelectric equation, which relates the energy of the incident photons to the work function of the metal and the kinetic energy of the emitted electrons.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The energy of a photon can be expressed as:
\[ E = h \nu \]
where \(h\) is Planck's constant and \(\nu\) is the frequency of the light.

2. **Setting Up the Equations**:
For the first frequency \(x\) with maximum kinetic energy \(y\):
\[ h x = \phi + y \quad \text{(Equation 1)} \]
where \(\phi\) is the work function of the metal.

For the second frequency \(2x\) with maximum kinetic energy \(3y\):
\[ h (2x) = \phi + 3y \quad \text{(Equation 2)} \]

3. **Substituting and Rearranging**:
From Equation 1, we can express \(\phi\):
\[ \phi = h x - y \]

Substitute this expression for \(\phi\) into Equation 2:
\[ h (2x) = (h x - y) + 3y \]

4. **Simplifying the Equation**:
This simplifies to:
\[ 2hx = h x + 2y \]
Rearranging gives:
\[ 2hx - hx = 2y \]
\[ hx = 2y \]

5. **Finding the Value of \(y\)**:
From the equation \(hx = 2y\), we can express \(y\) as:
\[ y = \frac{hx}{2} \]

6. **Substituting Back to Find \(\phi\)**:
Substitute \(y\) back into Equation 1:
\[ \phi = h x - \frac{hx}{2} \]
\[ \phi = \frac{hx}{2} \]

7. **Relating Work Function to Threshold Frequency**:
The work function \(\phi\) can also be expressed in terms of the threshold frequency \(\nu_0\):
\[ \phi = h \nu_0 \]
Setting these equal gives:
\[ h \nu_0 = \frac{hx}{2} \]

8. **Solving for Threshold Frequency**:
Dividing both sides by \(h\):
\[ \nu_0 = \frac{x}{2} \]

### Final Answer:
The threshold frequency of the metal is:
\[ \nu_0 = \frac{x}{2} \]