A particle of specific charge (charge/mass) `alpha` starts moving from the origin under the action of an electric field `oversetrarrE = E_(0)hati` and magnetic field
`oversetrarrB = B_(0)hatk` Its velocity at `(x_(0),y_(0).0)` is `(4hati +3hatj)` The value of `x_(0)` is .

Options

A. `(13)/(2)(alphaE_(0))/(B_(0))`
B. `(16alphaB_(0))/(E_(0))`
C. `(25)/(2alphaE_(0))`
D. `(5alpha)/(2B_(0))`

Answer

To solve the problem, we need to find the value of \( x_0 \) given the specific charge \( \alpha \), the electric field \( \vec{E} = E_0 \hat{i} \), the magnetic field \( \vec{B} = B_0 \hat{k} \), and the velocity \( \vec{v} = 4 \hat{i} + 3 \hat{j} \) at the point \( (x_0, y_0, 0) \).

### Step-by-Step Solution:

1. **Understanding Forces**:
The particle experiences an electric force due to the electric field and a magnetic force due to the magnetic field. The magnetic force does no work since it is always perpendicular to the direction of motion.

2. **Electric Force Calculation**:
The electric force \( \vec{F}_E \) acting on the particle is given by:
\[ \vec{F}_E = q \vec{E} = q E_0 \hat{i} \]
where \( q \) is the charge of the particle.

3. **Work Done by Electric Force**:
The work done \( W \) by the electric force as the particle moves from the origin to the point \( (x_0, y_0) \) is given by:
\[ W = \vec{F}_E \cdot \vec{d} \]
where \( \vec{d} = x_0 \hat{i} + y_0 \hat{j} \).

4. **Dot Product Calculation**:
The dot product \( \vec{F}_E \cdot \vec{d} \) becomes:
\[ W = (q E_0 \hat{i}) \cdot (x_0 \hat{i} + y_0 \hat{j}) = q E_0 x_0 \]

5. **Change in Kinetic Energy**:
The change in kinetic energy \( \Delta KE \) is given by:
\[ \Delta KE = KE_{final} - KE_{initial} \]
Since the particle starts from rest, \( KE_{initial} = 0 \). The final kinetic energy is:
\[ KE_{final} = \frac{1}{2} m v^2 \]
where \( v = \sqrt{(4)^2 + (3)^2} = 5 \). Thus:
\[ KE_{final} = \frac{1}{2} m (5^2) = \frac{25m}{2} \]

6. **Equating Work Done to Change in Kinetic Energy**:
Setting the work done equal to the change in kinetic energy:
\[ q E_0 x_0 = \frac{25m}{2} \]

7. **Substituting Specific Charge**:
The specific charge \( \alpha \) is defined as \( \alpha = \frac{q}{m} \). Thus, we can rewrite \( q \) as \( q = \alpha m \):
\[ (\alpha m) E_0 x_0 = \frac{25m}{2} \]

8. **Solving for \( x_0 \)**:
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ \alpha E_0 x_0 = \frac{25}{2} \]
Therefore:
\[ x_0 = \frac{25}{2 \alpha E_0} \]

### Final Result:
The value of \( x_0 \) is:
\[ x_0 = \frac{25}{2 \alpha E_0} \]