When a metal is illuminated with light of frequency f, the maximum kinetic energy of the photoelectrons is 1.2 eV. When the frequency is increased by 50% the maximum kinetic energy increases to 4.2 eV. What is the threshold frequency for this metal?

Explanation

To find the threshold frequency for the metal, we can use the photoelectric equation:

\[ KE = hf - \phi \]

where:
- \( KE \) is the maximum kinetic energy of the photoelectrons,
- \( h \) is Planck's constant,
- \( f \) is the frequency of the incident light,
- \( \phi \) is the work function of the metal, which is related to the threshold frequency \( f_0 \) by \( \phi = hf_0 \).

### Step 1: Write the equations for the two scenarios given in the problem.

1. For the first frequency \( f \):
\[ KE_1 = hf - \phi \quad \text{(1)} \]
Given \( KE_1 = 1.2 \, \text{eV} \).

2. For the second frequency \( f' \) which is increased by 50%:
\[ f' = 1.5f \]
Thus, the kinetic energy becomes:
\[ KE_2 = hf' - \phi \quad \text{(2)} \]
Given \( KE_2 = 4.2 \, \text{eV} \).

### Step 2: Substitute \( f' \) in equation (2).

Substituting \( f' \) into equation (2):
\[ 4.2 = h(1.5f) - \phi \]

### Step 3: Rewrite the equations.

Now we have two equations:
1. From equation (1):
\[ 1.2 = hf - \phi \quad \text{(3)} \]
2. From the modified equation (2):
\[ 4.2 = 1.5hf - \phi \quad \text{(4)} \]

### Step 4: Solve for \( \phi \) in terms of \( f \).

From equation (3):
\[ \phi = hf - 1.2 \quad \text{(5)} \]

### Step 5: Substitute \( \phi \) from (5) into (4).

Substituting (5) into (4):
\[ 4.2 = 1.5hf - (hf - 1.2) \]
Simplifying this:
\[ 4.2 = 1.5hf - hf + 1.2 \]
\[ 4.2 = 0.5hf + 1.2 \]

### Step 6: Solve for \( hf \).

Subtracting 1.2 from both sides:
\[ 4.2 - 1.2 = 0.5hf \]
\[ 3.0 = 0.5hf \]
\[ hf = \frac{3.0}{0.5} = 6.0 \, \text{eV} \quad \text{(6)} \]

### Step 7: Substitute \( hf \) back to find \( \phi \).

Now substituting \( hf = 6.0 \, \text{eV} \) back into equation (5):
\[ \phi = 6.0 - 1.2 = 4.8 \, \text{eV} \]

### Step 8: Find the threshold frequency \( f_0 \).

Using the relation \( \phi = hf_0 \):
\[ f_0 = \frac{\phi}{h} \]

Using \( h = 4.135667696 \times 10^{-15} \, \text{eV.s} \):
\[ f_0 = \frac{4.8}{4.135667696 \times 10^{-15}} \approx 1.16 \times 10^{15} \, \text{Hz} \]

### Final Answer:
The threshold frequency \( f_0 \) for this metal is approximately \( 1.16 \times 10^{15} \, \text{Hz} \).

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