The maximum velocity of the photoelectrons emitted from the surface is v when light of frequency n falls on a metal surface. If the incident frequency is increased to 3n, the maximum velocity of the ejected photoelectrons will be:

Options

A. less than ` sqrt3v`
B. v
C. more than `sqrt3v`
D. equal to sqrt3v`

Explanation

To solve the problem, we will use the photoelectric effect equation and analyze the changes in the maximum velocity of photoelectrons when the frequency of incident light is increased.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- When light of frequency \( n \) falls on a metal surface, the maximum velocity of the emitted photoelectrons is \( v \).
- According to the photoelectric effect, the kinetic energy (KE) of the emitted photoelectrons can be expressed as:
\[ KE = \frac{1}{2} m v^2 = h n - \phi \]
- Here, \( h \) is Planck's constant, \( n \) is the frequency of the incident light, and \( \phi \) is the work function of the metal.

2. **Forming the First Equation**:
- From the above expression, we can write:
\[ \frac{1}{2} m v^2 = h n - \phi \quad \text{(Equation 1)} \]

3. **Changing the Frequency**:
- Now, if the frequency of the incident light is increased to \( 3n \), we need to find the new maximum velocity of the emitted photoelectrons, denoted as \( v' \).
- The new kinetic energy can be expressed as:
\[ KE' = \frac{1}{2} m v'^2 = h (3n) - \phi \]

4. **Forming the Second Equation**:
- Thus, we can write:
\[ \frac{1}{2} m v'^2 = 3h n - \phi \quad \text{(Equation 2)} \]

5. **Dividing the Two Equations**:
- To find the relationship between \( v' \) and \( v \), we divide Equation 2 by Equation 1:
\[ \frac{\frac{1}{2} m v'^2}{\frac{1}{2} m v^2} = \frac{3h n - \phi}{h n - \phi} \]
- Simplifying this gives:
\[ \frac{v'^2}{v^2} = \frac{3h n - \phi}{h n - \phi} \]

6. **Rearranging the Equation**:
- Now we can express this as:
\[ v'^2 = v^2 \cdot \frac{3h n - \phi}{h n - \phi} \]

7. **Analyzing the Expression**:
- We notice that \( 3h n - \phi \) is greater than \( h n - \phi \) since \( 3n > n \). Therefore, the fraction \( \frac{3h n - \phi}{h n - \phi} \) is greater than 3.
- This implies:
\[ v'^2 > 3v^2 \]
- Taking the square root of both sides, we find:
\[ v' > \sqrt{3} v \]

8. **Conclusion**:
- The maximum velocity of the ejected photoelectrons when the frequency is increased to \( 3n \) will be greater than \( \sqrt{3} v \).

### Final Answer:
The maximum velocity of the ejected photoelectrons will be greater than \( \sqrt{3} v \).