An electron of mass m and charge e is accelerated by a potential difference V. It then enters a uniform magnetic field B applied perpendicular to its path. The radius of the circular path of the electron is

Options

A. `r=((2mV)/(eB^(2)))^(1//2)`
B. `r=((2meV)/(B^(2)))^(1//2)`
C. `r=((2mB)/(eV^(2)))^(1//2)`
D. `r=((2B^(2)V)/(em))^(1//2)`

Answer

To find the radius of the circular path of an electron accelerated by a potential difference \( V \) and then entering a uniform magnetic field \( B \) perpendicular to its path, we can follow these steps:

### Step 1: Determine the kinetic energy of the electron
When an electron is accelerated through a potential difference \( V \), it gains kinetic energy given by the equation:
\[ KE = eV \]
where \( e \) is the charge of the electron.

### Step 2: Relate kinetic energy to velocity
The kinetic energy can also be expressed in terms of the mass \( m \) of the electron and its velocity \( v \):
\[ KE = \frac{1}{2} mv^2 \]
Setting these two expressions for kinetic energy equal gives:
\[ eV = \frac{1}{2} mv^2 \]

### Step 3: Solve for the velocity \( v \)
Rearranging the equation to solve for \( v \):
\[ mv^2 = 2eV \implies v^2 = \frac{2eV}{m} \implies v = \sqrt{\frac{2eV}{m}} \]

### Step 4: Use the formula for the radius of the circular path
When a charged particle moves in a magnetic field, the radius \( r \) of its circular path is given by:
\[ r = \frac{mv}{qB} \]
where \( q \) is the charge of the particle (in this case, \( e \) for the electron).

### Step 5: Substitute the expression for \( v \) into the radius formula
Substituting \( v \) from Step 3 into the radius formula:
\[ r = \frac{m \sqrt{\frac{2eV}{m}}}{eB} \]

### Step 6: Simplify the expression for \( r \)
This simplifies to:
\[ r = \frac{m \sqrt{2eV}}{eB\sqrt{m}} = \frac{\sqrt{2m} \sqrt{eV}}{eB} \]
Thus, we can express the radius as:
\[ r = \sqrt{\frac{2mV}{eB^2}} \]

### Final Expression
The final expression for the radius of the circular path of the electron is:
\[ r = \sqrt{\frac{2mV}{eB^2}} \]