The maximum velocity of electrons emitted from a metal surface is v. What would be the maximum velocity if the frequency of incident lightis increased by a factor of 4?

Options

A. more than `sqrt(3)` v
B. equal to `sqrt(3)` v
C. v
D. less than `sqrt(3)` v

Answer

To solve the problem, we need to apply the principles of the photoelectric effect, specifically Einstein's equation for the kinetic energy of emitted electrons.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
According to Einstein's equation, the maximum kinetic energy (K.E.) of the emitted electrons can be expressed as:
\[ K.E. = hf - \phi \]
where \( h \) is Planck's constant, \( f \) is the frequency of the incident light, and \( \phi \) is the work function of the metal.

2. **Initial Condition**:
Let the initial frequency of the incident light be \( f \). The maximum kinetic energy of the emitted electrons is given as:
\[ K.E. = \frac{1}{2}mv^2 = hf - \phi \]
where \( v \) is the maximum velocity of the emitted electrons.

3. **New Condition with Increased Frequency**:
Now, if the frequency of the incident light is increased by a factor of 4, the new frequency becomes:
\[ f' = 4f \]
The new maximum kinetic energy can be expressed as:
\[ K.E.' = hf' - \phi = h(4f) - \phi = 4hf - \phi \]

4. **Relating the Kinetic Energies**:
For the new maximum velocity \( v' \), we can write:
\[ K.E.' = \frac{1}{2}mv'^2 = 4hf - \phi \]

5. **Setting Up the Equations**:
We now have two equations:
- For the initial condition:
\[ \frac{1}{2}mv^2 = hf - \phi \quad \text{(1)} \]
- For the new condition:
\[ \frac{1}{2}mv'^2 = 4hf - \phi \quad \text{(2)} \]

6. **Dividing the Equations**:
To find the relationship between \( v' \) and \( v \), we can divide equation (2) by equation (1):
\[ \frac{mv'^2}{mv^2} = \frac{4hf - \phi}{hf - \phi} \]
Simplifying gives:
\[ \frac{v'^2}{v^2} = \frac{4hf - \phi}{hf - \phi} \]

7. **Finding the Maximum Velocity**:
Let’s denote \( C = \frac{\phi}{hf} \). Then we can rewrite the equation:
\[ \frac{v'^2}{v^2} = \frac{4 - C}{1 - C} \]
Taking the square root on both sides gives:
\[ \frac{v'}{v} = \sqrt{\frac{4 - C}{1 - C}} \]
Thus, we can express \( v' \) as:
\[ v' = v \sqrt{\frac{4 - C}{1 - C}} \]

8. **Conclusion**:
The maximum velocity of the emitted electrons increases when the frequency of the incident light is increased by a factor of 4. The exact relationship depends on the work function of the metal, but it is clear that \( v' \) will be greater than \( v \).