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 need to analyze the relationship between the frequency of incident light and the maximum velocity of photoelectrons emitted from a metal surface. We will use the photoelectric effect principles and the kinetic energy of the emitted electrons.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions:**
- Let the initial frequency of the light be \( n \).
- The maximum velocity of the emitted photoelectrons at this frequency is \( v \).
- The kinetic energy (KE) of the emitted photoelectrons can be expressed as:
\[ KE = \frac{1}{2}mv^2 = nh - \phi \]
where \( h \) is Planck's constant and \( \phi \) is the work function of the metal.

2. **Setting Up the Equation for Initial Conditions:**
- From the equation above, we can rearrange it to express \( nh \):
\[ nh = \frac{1}{2}mv^2 + \phi \quad \text{(Equation 1)} \]

3. **Analyzing the New Conditions:**
- Now, when the frequency of the light is increased to \( 3n \), the new kinetic energy of the emitted photoelectrons becomes:
\[ KE' = \frac{1}{2}mV^2 = 3nh - \phi \]
where \( V \) is the new maximum velocity of the emitted photoelectrons.

4. **Setting Up the Equation for New Conditions:**
- Rearranging the new kinetic energy equation gives:
\[ 3nh = \frac{1}{2}mV^2 + \phi \quad \text{(Equation 2)} \]

5. **Substituting Equation 1 into Equation 2:**
- We can substitute \( nh \) from Equation 1 into Equation 2:
\[ 3\left(\frac{1}{2}mv^2 + \phi\right) = \frac{1}{2}mV^2 + \phi \]
- Expanding this gives:
\[ \frac{3}{2}mv^2 + 3\phi = \frac{1}{2}mV^2 + \phi \]

6. **Rearranging the Equation:**
- Rearranging the terms leads to:
\[ \frac{3}{2}mv^2 + 2\phi = \frac{1}{2}mV^2 \]
- Multiplying through by 2 to eliminate the fraction:
\[ 3mv^2 + 4\phi = mV^2 \]

7. **Solving for \( V^2 \):**
- Rearranging gives:
\[ mV^2 = 3mv^2 + 4\phi \]
- Dividing by \( m \):
\[ V^2 = 3v^2 + \frac{4\phi}{m} \]

8. **Finding the Relationship Between \( V \) and \( v \):**
- Since \( \frac{4\phi}{m} \) is a positive term, we can conclude:
\[ V^2 > 3v^2 \]
- Taking the square root:
\[ V > \sqrt{3}v \]

### Conclusion:
Thus, the maximum velocity of the ejected photoelectrons when the frequency is increased to \( 3n \) is greater than \( \sqrt{3}v \).

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

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