In Bainbridge mass spectograph a potential difference of 1000 V is applied between two plates distant 1 cm apart and magnetic field in `B = 1T`. The velocity of unflected positive ions in m//s from the velocity selector is

Options

A. `10^(7) m//s`
B. `10^(4) m//s`
C. `10^(5) m//s`
D. `10^(2) m//s`

Answer

To find the velocity of the undeflected positive ions in a Bainbridge mass spectrograph, we can follow these steps:

### Step 1: Understand the relationship between electric and magnetic forces
In the Bainbridge mass spectrograph, the electric force (Fe) acting on the ions is equal to the magnetic force (Fb) when the ions are undeflected. This can be expressed as:
\[ F_e = F_b \]
Where:
- \( F_e = qE \) (Electric force)
- \( F_b = qvB \) (Magnetic force)

### Step 2: Set the forces equal to each other
Since the forces are equal, we can write:
\[ qE = qvB \]
We can cancel the charge \( q \) from both sides (assuming \( q \neq 0 \)):
\[ E = vB \]

### Step 3: Calculate the electric field (E)
The electric field (E) between the plates can be calculated using the potential difference (V) and the distance (d) between the plates:
\[ E = \frac{V}{d} \]
Given:
- \( V = 1000 \, \text{V} \)
- \( d = 1 \, \text{cm} = 0.01 \, \text{m} \)

Substituting the values:
\[ E = \frac{1000 \, \text{V}}{0.01 \, \text{m}} = 100000 \, \text{V/m} = 10^5 \, \text{V/m} \]

### Step 4: Substitute E into the equation to find velocity (v)
Now, we can substitute \( E \) into the equation \( E = vB \):
\[ 10^5 = v \cdot 1 \]
Thus,
\[ v = 10^5 \, \text{m/s} \]

### Final Answer
The velocity of the undeflected positive ions is:
\[ v = 100000 \, \text{m/s} \]

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