In Thomson mass spectrograph `vecE bot vecB` then the velocity of underflected electron beam will be :

Options

A. `(|vecE|)/(|vecB|)`
B. `vecExxvecB`
C. `(|vecB|)/(|vecE|)`
D. `E^2/B^2`

Answer

To solve the problem, we need to determine the velocity of an electron beam in a Thomson mass spectrograph when the electric field (\( \vec{E} \)) is perpendicular to the magnetic field (\( \vec{B} \)). The condition for the electron beam to be unaffected by the fields is that the forces acting on the electrons due to the electric and magnetic fields must be equal and opposite.

### Step-by-Step Solution:

1. **Identify the Forces**:
- The force acting on a charge \( Q \) in an electric field \( \vec{E} \) is given by:
\[ F_E = Q \vec{E} \]
- The force acting on a charge \( Q \) moving with velocity \( \vec{v} \) in a magnetic field \( \vec{B} \) is given by:
\[ F_B = Q (\vec{v} \times \vec{B}) \]

2. **Set the Forces Equal**:
For the electron beam to be unaffected, these two forces must be equal in magnitude:
\[ F_E = F_B \]
This leads to the equation:
\[ Q \vec{E} = Q (\vec{v} \times \vec{B}) \]

3. **Cancel the Charge**:
Since the charge \( Q \) is the same on both sides, we can cancel it out (assuming \( Q \neq 0 \)):
\[ \vec{E} = \vec{v} \times \vec{B} \]

4. **Magnitude of the Velocity**:
Since \( \vec{E} \) and \( \vec{B} \) are perpendicular, we can express the magnitude of the velocity \( v \) as:
\[ E = vB \]
Rearranging this gives:
\[ v = \frac{E}{B} \]

5. **Final Expression**:
Therefore, the velocity of the underreflected electron beam is:
\[ v = \frac{E}{B} \]

### Conclusion:
The velocity of the underreflected electron beam in the Thomson mass spectrograph when the electric field is perpendicular to the magnetic field is given by:
\[ v = \frac{E}{B} \]