In a photoelectric effect experiment, for radiation with frequency `v_0` with `hv_0` = 8eV, electrons are emitted with energy 2 eV. What is the energy of the electrons emitted for incoming radiation of frequency 1.25 `v_0`?

Options

A. 1eV
B. 3.25 eV
C. 4eV
D. 9,25 eV

Explanation

To solve the problem step by step, we will follow the principles of the photoelectric effect.

### Step 1: Understand the given information
We are given:
- The energy of the incoming radiation at frequency \( v_0 \) is \( hv_0 = 8 \, \text{eV} \).
- The kinetic energy of the emitted electrons at this frequency is \( KE = 2 \, \text{eV} \).

### Step 2: Calculate the work function
The work function \( \phi \) of the metal can be calculated using the formula:
\[ \phi = hv_0 - KE \]
Substituting the values:
\[ \phi = 8 \, \text{eV} - 2 \, \text{eV} = 6 \, \text{eV} \]

### Step 3: Determine the energy for incoming radiation of frequency \( 1.25 v_0 \)
The energy of the incoming radiation at frequency \( 1.25 v_0 \) can be calculated as:
\[ E = h(1.25 v_0) = 1.25 \times hv_0 \]
Substituting \( hv_0 = 8 \, \text{eV} \):
\[ E = 1.25 \times 8 \, \text{eV} = 10 \, \text{eV} \]

### Step 4: Calculate the maximum kinetic energy of emitted electrons
The maximum kinetic energy of the emitted electrons when the incoming radiation energy is \( 10 \, \text{eV} \) can be calculated as:
\[ KE_{max} = E - \phi \]
Substituting the values:
\[ KE_{max} = 10 \, \text{eV} - 6 \, \text{eV} = 4 \, \text{eV} \]

### Final Answer
The energy of the electrons emitted for incoming radiation of frequency \( 1.25 v_0 \) is \( 4 \, \text{eV} \).

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