What is the speed acquired by an electron when it is accelerated from rest through 1 V potential difference?

To find the speed acquired by an electron when it is accelerated from rest through a potential difference of 1 volt, we can use the relationship between electric potential energy and kinetic energy. When an electron is accelerated through a potential difference, it gains kinetic energy equal to the work done on it by the electric field.

Understanding the Energy Conversion

When an electron is accelerated through a potential difference (V), it gains kinetic energy (KE) given by the equation:

• KE = qV

Here, q is the charge of the electron, and V is the potential difference. The charge of an electron is approximately 1.6 x 10^-19 coulombs.

Calculating the Kinetic Energy

For a potential difference of 1 volt, the kinetic energy gained by the electron can be calculated as follows:

• KE = (1.6 x 10^-19 C) * (1 V) = 1.6 x 10^-19 joules

Relating Kinetic Energy to Speed

The kinetic energy of an object can also be expressed in terms of its mass and speed:

• KE = (1/2)mv²

In this equation, m is the mass of the electron, which is approximately 9.11 x 10^-31 kg, and v is the speed we want to find.

Setting the Equations Equal

Now we can set the two expressions for kinetic energy equal to each other:

• (1/2)mv² = qV

Substituting the known values into the equation:

• (1/2)(9.11 x 10^-31 kg)v² = 1.6 x 10^-19 joules

Solving for Speed

Now, we can solve for v:

• v² = (2 * 1.6 x 10^-19) / (9.11 x 10^-31)
• v² = (3.2 x 10^-19) / (9.11 x 10^-31)
• v² ≈ 3.51 x 10¹¹
• v ≈ √(3.51 x 10¹¹)
• v ≈ 1.87 x 10⁶ m/s

Final Result

Thus, the speed acquired by the electron when it is accelerated through a potential difference of 1 volt is approximately 1.87 x 10⁶ meters per second. This speed is quite significant, illustrating how even a small potential difference can impart a considerable amount of energy to a charged particle like an electron.