Lights of two different frequencies whose photons have energies 1 and 2.5 eV, respectively, successively illuminate a metal whose work function is 0.5 eV. The ratio of the maximum speeds of the emitted electrons

Options

A. `1:2`
B. `1:1`
C. `1:5`
D. `1:4`

Explanation

To solve the problem, we will use the photoelectric effect equation, which relates the energy of the incoming photons to the work function of the metal and the kinetic energy of the emitted electrons.

### Step-by-Step Solution:

1. **Understand the photoelectric equation**:
The photoelectric equation is given by:
\[ E = \phi + \frac{1}{2} mv^2 \]
where \(E\) is the energy of the incoming photon, \(\phi\) is the work function of the metal, \(m\) is the mass of the electron, and \(v\) is the speed of the emitted electron.

2. **Rearranging the equation**:
Rearranging the equation to find the kinetic energy of the emitted electrons gives:
\[ \frac{1}{2} mv^2 = E - \phi \]

3. **Define energies for the two photons**:
Let:
- \(E_1 = 1 \, \text{eV}\) (energy of the first photon)
- \(E_2 = 2.5 \, \text{eV}\) (energy of the second photon)
- \(\phi = 0.5 \, \text{eV}\) (work function of the metal)

4. **Write the equations for the two emitted electrons**:
For the first photon:
\[ \frac{1}{2} mv_1^2 = E_1 - \phi = 1 - 0.5 = 0.5 \, \text{eV} \]
For the second photon:
\[ \frac{1}{2} mv_2^2 = E_2 - \phi = 2.5 - 0.5 = 2.0 \, \text{eV} \]

5. **Express the maximum speeds**:
From the equations, we can express the speeds:
\[ v_1^2 = \frac{2(E_1 - \phi)}{m} = \frac{2 \times 0.5}{m} = \frac{1}{m} \]
\[ v_2^2 = \frac{2(E_2 - \phi)}{m} = \frac{2 \times 2.0}{m} = \frac{4}{m} \]

6. **Find the ratio of speeds**:
The ratio of the squares of the speeds is:
\[ \frac{v_1^2}{v_2^2} = \frac{\frac{1}{m}}{\frac{4}{m}} = \frac{1}{4} \]
Therefore, taking the square root to find the ratio of the speeds:
\[ \frac{v_1}{v_2} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

### Final Answer:
The ratio of the maximum speeds of the emitted electrons \(v_1 : v_2\) is \(1 : 2\).