Initially, energy of incident photon was double of the work function of metal in a photoelectric effect. Finally energy of incident photon made 10 times of the work function. The ratio of maximum possible speed of photo electrons in initial & final case is `alpha/beta`. Then find minimum value of `alpha` :

Options

A. 2
B. 1
C. 4
D. 3

Answer

To solve the problem, we will use the principles of the photoelectric effect. We will calculate the maximum kinetic energy of the emitted electrons in both cases and then find the ratio of their speeds.

### Step 1: Define the energies in both cases

1. **Initial Case (Case 1)**:
- The energy of the incident photon \( E_1 = 2\phi \) (where \( \phi \) is the work function).
- The kinetic energy of the emitted electron in this case is given by:
\[ KE_1 = E_1 - \phi = 2\phi - \phi = \phi \]

2. **Final Case (Case 2)**:
- The energy of the incident photon \( E_2 = 10\phi \).
- The kinetic energy of the emitted electron in this case is given by:
\[ KE_2 = E_2 - \phi = 10\phi - \phi = 9\phi \]

### Step 2: Relate kinetic energy to speed

The kinetic energy of an electron is related to its speed \( v \) by the formula:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the electron.

From this, we can express the speed in terms of kinetic energy:
\[ v = \sqrt{\frac{2 \cdot KE}{m}} \]

### Step 3: Calculate speeds in both cases

1. **Speed in Initial Case (Case 1)**:
\[ v_1 = \sqrt{\frac{2 \cdot KE_1}{m}} = \sqrt{\frac{2 \cdot \phi}{m}} \]

2. **Speed in Final Case (Case 2)**:
\[ v_2 = \sqrt{\frac{2 \cdot KE_2}{m}} = \sqrt{\frac{2 \cdot 9\phi}{m}} = \sqrt{\frac{18\phi}{m}} \]

### Step 4: Find the ratio of speeds

Now, we find the ratio of the maximum possible speeds of photoelectrons in the initial and final cases:
\[ \frac{v_1}{v_2} = \frac{\sqrt{\frac{2\phi}{m}}}{\sqrt{\frac{18\phi}{m}}} = \frac{\sqrt{2\phi}}{\sqrt{18\phi}} = \frac{\sqrt{2}}{\sqrt{18}} = \frac{\sqrt{2}}{3\sqrt{2}} = \frac{1}{3} \]

### Step 5: Express the ratio in terms of \( \alpha \) and \( \beta \)

The problem states that the ratio of maximum possible speed of photoelectrons in the initial and final cases is \( \frac{\alpha}{\beta} \). From our calculations, we have:
\[ \frac{v_1}{v_2} = \frac{1}{3} \implies \frac{\alpha}{\beta} = \frac{1}{3} \]

This means \( \alpha = 1 \) and \( \beta = 3 \).

### Conclusion

The minimum value of \( \alpha \) is:
\[ \boxed{1} \]