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. 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 will use the photoelectric effect principles and the relationship between the energy of the incident photons, the work function of the metal, and the kinetic energy of the emitted photoelectrons.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The maximum kinetic energy (KE) of the emitted photoelectrons can be expressed using the equation:
\[ KE = E_{\text{photon}} - \phi \]
where \(E_{\text{photon}} = h \nu\) is the energy of the incident photon, \(h\) is Planck's constant, \(\nu\) is the frequency of the incident light, and \(\phi\) is the work function of the metal.

2. **For the first scenario (frequency = \(n\))**:
The maximum kinetic energy of the photoelectrons when light of frequency \(n\) falls on the metal surface is:
\[ KE_1 = h n - \phi \]
The maximum velocity \(v\) of the emitted photoelectrons is related to the kinetic energy by:
\[ KE_1 = \frac{1}{2} m v^2 \]
Therefore, we can write:
\[ h n - \phi = \frac{1}{2} m v^2 \quad \text{(1)} \]

3. **For the second scenario (frequency = \(3n\))**:
When the frequency is increased to \(3n\), the maximum kinetic energy becomes:
\[ KE_2 = h (3n) - \phi = 3h n - \phi \]
Let the new maximum velocity of the emitted photoelectrons be \(v'\). Then we have:
\[ KE_2 = \frac{1}{2} m (v')^2 \quad \text{(2)} \]

4. **Setting up the equations**:
From equation (1):
\[ h n - \phi = \frac{1}{2} m v^2 \]
From equation (2):
\[ 3h n - \phi = \frac{1}{2} m (v')^2 \]

5. **Subtracting the two equations**:
To find the relationship between \(v\) and \(v'\), we can subtract equation (1) from equation (2):
\[ (3h n - \phi) - (h n - \phi) = \frac{1}{2} m (v')^2 - \frac{1}{2} m v^2 \]
Simplifying gives:
\[ 2h n = \frac{1}{2} m (v')^2 - \frac{1}{2} m v^2 \]
\[ 2h n = \frac{1}{2} m \left( (v')^2 - v^2 \right) \]
Multiplying through by 2:
\[ 4h n = m \left( (v')^2 - v^2 \right) \]

6. **Expressing \(v'\)**:
Rearranging gives:
\[ (v')^2 = v^2 + \frac{4h n}{m} \]
Taking the square root:
\[ v' = \sqrt{v^2 + \frac{4h n}{m}} \]

7. **Finding the relationship**:
Since \(h n = \frac{1}{2} m v^2 + \phi\), we can express \(\frac{4h n}{m}\) in terms of \(v^2\):
\[ v' = \sqrt{v^2 + 2v^2} = \sqrt{3v^2 + \frac{4h n}{m}} \]
Since \(\frac{4h n}{m}\) is a positive term, we conclude that:
\[ v' > \sqrt{3}v \]

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