The maximum velocity of electrons emitted from a metal surface is v when the frequecny of light falling on it is f. The maximum velocity when the frequency becomes 4f is

Options

A. 2v
B. `gt 2v`
C. `lt2v`
D. between 2v and 4v

Answer

To solve the problem, we need to analyze the relationship between the frequency of light incident on a metal surface and the maximum velocity of the emitted electrons. This is based on the photoelectric effect.

### Step-by-Step Solution:

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

2. **Expressing Kinetic Energy in Terms of Velocity**:
The kinetic energy can also be expressed in terms of the maximum velocity \( v \) of the emitted electrons:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the electron.

3. **Setting Up the First Condition**:
For the first case, where the frequency is \( f \) and the maximum velocity is \( v \):
\[ \frac{1}{2} mv^2 = hf - \phi \]

4. **Setting Up the Second Condition**:
Now, when the frequency becomes \( 4f \), we denote the new maximum velocity as \( v' \):
\[ \frac{1}{2} mv'^2 = h(4f) - \phi \]
Simplifying this gives:
\[ \frac{1}{2} mv'^2 = 4hf - \phi \]

5. **Relating the Two Conditions**:
We can express the second equation in terms of the first:
\[ \frac{1}{2} mv'^2 = 4hf - \phi = 4(hf - \phi) + 3\phi \]
Substituting the expression for \( hf - \phi \) from the first condition:
\[ \frac{1}{2} mv'^2 = 4 \left( \frac{1}{2} mv^2 \right) + 3\phi \]
This simplifies to:
\[ mv'^2 = 4mv^2 + 6\phi \]

6. **Finding the New Velocity**:
Dividing through by \( m \):
\[ v'^2 = 4v^2 + \frac{6\phi}{m} \]
Taking the square root gives:
\[ v' = \sqrt{4v^2 + \frac{6\phi}{m}} \]

7. **Analyzing the Result**:
Since \( \frac{6\phi}{m} \) is a positive term, we can conclude that:
\[ v' > \sqrt{4v^2} = 2v \]
Therefore, the maximum velocity when the frequency becomes \( 4f \) is greater than \( 2v \).

### Final Answer:
The maximum velocity when the frequency becomes \( 4f \) is greater than \( 2v \).