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 effect 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 Equation**:
The energy of the incident light can be expressed as:
\[ E = h \nu = \phi + KE_{\text{max}} \]
where \(E\) is the energy of the incident photon, \(h\) is Planck's constant, \(\nu\) is the frequency of the light, \(\phi\) is the work function of the metal, and \(KE_{\text{max}}\) is the maximum kinetic energy of the ejected electrons.

2. **Setting Up the Equations**:
For the first light radiation with frequency \(x\) and maximum kinetic energy \(y\):
\[ h x = \phi + y \tag{1} \]
For the second light radiation with frequency \(2x\) and maximum kinetic energy \(3y\):
\[ h (2x) = \phi + 3y \tag{2} \]

3. **Subtracting the Equations**:
Now, we will subtract equation (1) from equation (2):
\[ h(2x) - h(x) = (\phi + 3y) - (\phi + y) \]
This simplifies to:
\[ h(2x - x) = 3y - y \]
\[ h x = 2y \tag{3} \]

4. **Expressing \(y\) in Terms of \(x\)**:
From equation (3), we can express \(y\):
\[ y = \frac{h x}{2} \tag{4} \]

5. **Substituting \(y\) Back into Equation (1)**:
Now we substitute equation (4) back into equation (1):
\[ h x = \phi + \frac{h x}{2} \]
Rearranging gives:
\[ \phi = h x - \frac{h x}{2} \]
\[ \phi = \frac{h x}{2} \tag{5} \]

6. **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 equation (5) equal to this gives:
\[ h \nu_0 = \frac{h x}{2} \]

7. **Solving for Threshold Frequency**:
Dividing both sides by \(h\) (assuming \(h \neq 0\)):
\[ \nu_0 = \frac{x}{2} \]

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