The threshold frequency for a certain photosensitive metal is `v_(0)`. When it is illuminated by light of frequency `v=2 v_(0)`, the maximum velocity of photoelectrons is `v_(0)`.The maximum velocity of the photoelectrons when the same metal is illuminated by light of frequency `v=5 v_(0)` is `x v_(0)`. Find `x`

Answer

To solve the problem, we will use the concepts of the photoelectric effect, specifically the relationship between the frequency of incident light, the threshold frequency, and the kinetic energy of emitted photoelectrons.

### Step-by-Step Solution:

1. **Understanding the Threshold Frequency**:
The threshold frequency \( v_0 \) is the minimum frequency of light required to eject electrons from the metal surface. For any frequency \( v \) greater than \( v_0 \), photoelectrons will be emitted.

2. **Kinetic Energy of Photoelectrons**:
The kinetic energy (KE) of the emitted photoelectrons can be expressed using the equation:
\[ KE = h(v - v_0) \]
where \( h \) is Planck's constant, \( v \) is the frequency of the incident light, and \( v_0 \) is the threshold frequency.

3. **Case 1: Frequency \( v = 2v_0 \)**:
- When the frequency of the light is \( v = 2v_0 \), the kinetic energy of the emitted photoelectrons can be calculated as:
\[ KE = h(2v_0 - v_0) = h(v_0) \]
- The maximum velocity \( v_{max} \) of the photoelectrons is given as \( v_0 \). The kinetic energy can also be expressed in terms of velocity:
\[ KE = \frac{1}{2} mv_{max}^2 \]
- Substituting \( v_{max} = v_0 \):
\[ h(v_0) = \frac{1}{2} m (v_0)^2 \quad \text{(1)} \]

4. **Case 2: Frequency \( v = 5v_0 \)**:
- Now, we consider the case when the frequency of light is \( v = 5v_0 \):
\[ KE = h(5v_0 - v_0) = h(4v_0) \]
- Let the maximum velocity of the photoelectrons in this case be \( v_{max}' \). The kinetic energy can again be expressed as:
\[ KE = \frac{1}{2} mv_{max}'^2 \]
- Thus, we have:
\[ h(4v_0) = \frac{1}{2} mv_{max}'^2 \quad \text{(2)} \]

5. **Relating Equations (1) and (2)**:
- From equation (1), we can express \( h \) in terms of \( v_0 \):
\[ h = \frac{1}{2} m \frac{(v_0)^2}{v_0} = \frac{1}{2} mv_0 \]
- Substituting this into equation (2):
\[ \frac{1}{2} mv_0 (4v_0) = \frac{1}{2} mv_{max}'^2 \]
- Simplifying gives:
\[ 4v_0^2 = v_{max}'^2 \]

6. **Finding Maximum Velocity**:
- Taking the square root of both sides:
\[ v_{max}' = 2v_0 \]

7. **Finding \( x \)**:
- Since we are asked for the maximum velocity in the form \( x v_0 \), we have:
\[ v_{max}' = 2v_0 \implies x = 2 \]

### Final Answer:
Thus, the value of \( x \) is \( 2 \).