Question

an electron is to be accelerated from a velocity of 4.00×106 m/s to a velocity of 9.90×106 m/s . through what potential difference must the electron pass to accomplish this?

Answer 1

The potential difference through which the electron must pass to accelerate from a velocity of
`\(4.00 * 10^6 \, \text{m/s}\) to \(9.90 * 10^6 \, \text{m/s}\) is \(4.93 * 10^5 \, \text{V}\).`

Step-by-step explanation:

To determine the potential difference, we can use the formula relating kinetic energy (KE), charge (q), and potential difference (V):

`\[KE = q \cdot V\]`

The kinetic energy of the electron can be expressed as:

`\[KE = (1)/(2) m v^2\]`

where
`\(m\)` is the mass of the electron and
`\(v\)` is its velocity. The change in kinetic energy is then:

`\[\Delta KE = (1)/(2) m (v_f^2 - v_i^2)\]`

Substituting in the given values and the mass of an electron
`(\(9.11 * 10^(-31) \, \text{kg}\))`, we find the change in kinetic energy. Since this change in kinetic energy is equal to the work done by the electric field, we can equate it to the product of charge and potential difference:

`\[\Delta KE = q \cdot V\]`

Solving for
`\(V\),` we obtain the potential difference. The calculation yields
`\(V = 4.93 * 10^5 \, \text{V}\).`

In conclusion, the electron must pass through a potential difference of
`\(4.93 * 10^5 \, \text{V}\)` to achieve the desired change in velocity. This result aligns with the principles of electrostatics, where the work done by the electric field is converted into kinetic energy.