In an electron gun, electron are accelerated through a potential difference of V volt. Taking electronic charge and mass to be respectively 'e' and 'm', the maximum velocity attained by them is

Options

A. `(2eV)/(m)`
B. `sqrt((2eV)/(m))`
C. `(2m)/(eV)`
D. `(V^(2))/(2em)`

Answer

To find the maximum velocity attained by electrons accelerated through a potential difference \( V \), we can follow these steps:

### Step 1: Understand the relationship between work done and kinetic energy
When an electron is accelerated through a potential difference \( V \), the work done on the electron is converted into kinetic energy. The work done \( W \) on the electron can be expressed as:
\[ W = eV \]
where \( e \) is the charge of the electron.

### Step 2: Relate work done to kinetic energy
The kinetic energy \( KE \) of the electron can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the electron and \( v \) is its velocity.

### Step 3: Set the work done equal to the kinetic energy
Since the work done on the electron is equal to the kinetic energy gained, we can set the two equations equal to each other:
\[ eV = \frac{1}{2} mv^2 \]

### Step 4: Solve for maximum velocity \( v \)
To find the maximum velocity \( v \), we rearrange the equation:
\[ mv^2 = 2eV \]
\[ v^2 = \frac{2eV}{m} \]
Taking the square root of both sides gives us:
\[ v = \sqrt{\frac{2eV}{m}} \]

### Conclusion
Thus, the maximum velocity attained by the electrons is:
\[ v = \sqrt{\frac{2eV}{m}} \]