When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is:

Options

A. 0.65 eV
B. 1.0 eV
C. 1.3 eV
D. 1.5eV

Explanation

To solve the problem, we need to determine the work function (φ) of the metal based on the given information about the kinetic energy of the emitted photoelectrons and the change in energy of the incident radiation.

### Step-by-Step Solution:

1. **Define the initial conditions:**
Let the initial energy of the incident radiation be \( E_1 = h\nu_1 \). The kinetic energy of the emitted photoelectrons at this point is given as:
\[ KE_1 = 0.5 \, \text{eV} \]
According to the photoelectric equation:
\[ h\nu_1 = KE_1 + \phi \]
Therefore, we can write:
\[ h\nu_1 = 0.5 + \phi \quad \text{(1)} \]

2. **Define the new conditions after increasing the energy:**
The energy of the incident radiation is increased by 20%, so the new energy is:
\[ E_2 = h\nu_2 = 1.2 \times h\nu_1 \]
The new kinetic energy of the emitted photoelectrons is:
\[ KE_2 = 0.8 \, \text{eV} \]
Again, using the photoelectric equation:
\[ h\nu_2 = KE_2 + \phi \]
Substituting the values, we have:
\[ 1.2 \times h\nu_1 = 0.8 + \phi \quad \text{(2)} \]

3. **Set up the equations:**
Now we have two equations:
- From equation (1): \( h\nu_1 = 0.5 + \phi \)
- From equation (2): \( 1.2 \times h\nu_1 = 0.8 + \phi \)

4. **Substitute equation (1) into equation (2):**
Substitute \( h\nu_1 \) from equation (1) into equation (2):
\[ 1.2 \times (0.5 + \phi) = 0.8 + \phi \]

5. **Expand and simplify:**
Expanding the left side:
\[ 0.6 + 1.2\phi = 0.8 + \phi \]
Rearranging gives:
\[ 1.2\phi - \phi = 0.8 - 0.6 \]
\[ 0.2\phi = 0.2 \]

6. **Solve for φ:**
Dividing both sides by 0.2:
\[ \phi = 1 \, \text{eV} \]

### Final Answer:
The work function of the metal is \( \phi = 1 \, \text{eV} \).