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 solve the problem, we will use the photoelectric equation, which relates the maximum kinetic energy (K.E.) of the emitted photoelectrons to the frequency of the incident light and the threshold frequency of the metal.

The photoelectric equation is given by:

\[ K.E. = hf - \phi \]

where:
- \( K.E. \) 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 can also be expressed in terms of the threshold frequency \( f_0 \) as \( \phi = hf_0 \).

### Step 1: Write the equations for the two scenarios

1. For the first scenario (frequency \( f \)):
\[ K.E. = hf - \phi \implies 1.2 \text{ eV} = hf - hf_0 \tag{1} \]

2. For the second scenario (frequency increased by 50%, which means frequency is \( 1.5f \)):
\[ K.E. = hf' - \phi \implies 4.2 \text{ eV} = h(1.5f) - hf_0 \tag{2} \]

### Step 2: Simplify both equations

From equation (1):
\[ hf - hf_0 = 1.2 \implies hf = 1.2 + hf_0 \tag{3} \]

From equation (2):
\[ h(1.5f) - hf_0 = 4.2 \implies 1.5hf - hf_0 = 4.2 \tag{4} \]

### Step 3: Substitute equation (3) into equation (4)

Substituting \( hf \) from equation (3) into equation (4):
\[ 1.5(1.2 + hf_0) - hf_0 = 4.2 \]

### Step 4: Expand and simplify

Expanding the left side:
\[ 1.8 + 1.5hf_0 - hf_0 = 4.2 \]
\[ 1.8 + 0.5hf_0 = 4.2 \]

### Step 5: Solve for \( hf_0 \)

Subtract 1.8 from both sides:
\[ 0.5hf_0 = 4.2 - 1.8 \]
\[ 0.5hf_0 = 2.4 \]

Now, multiply both sides by 2:
\[ hf_0 = 4.8 \]

### Step 6: Find the threshold frequency \( f_0 \)

To find \( f_0 \), we use the relationship \( hf_0 = \phi \):
\[ f_0 = \frac{4.8}{h} \]

### Conclusion

The threshold frequency \( f_0 \) can be expressed in terms of the known constants. However, since we are asked for the value in terms of electronvolts, we can conclude that:

\[ \phi = 4.8 \text{ eV} \implies f_0 = \frac{4.8 \text{ eV}}{h} \]

### Final Answer

The threshold frequency \( f_0 \) for this metal is \( 0.8f \) or \( \frac{4.8}{h} \).

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