An electron has an initial speed of 4.00×106 m/s. What potential difference (sign and magnitude) is required to bring the electron to rest?
To solve this problem, we need to understand the relationship between the kinetic energy of an electron and the potential difference it encounters.
First, let's recall the kinetic energy formula for an object:
KE = 1/2 * m * v²
where KE represents the kinetic energy, m is the mass of the object, and v is its velocity.
In the case of an electron, its mass (m) is approximately 9.11 × 10^-31 kg, and its initial velocity (v) is 4.00 × 10^6 m/s.
Now, let's consider the relationship between potential difference (V) and kinetic energy (KE) for an electron.
The potential energy (PE) an electron gains or loses when moving through a potential difference is given by:
PE = q * V
where q is the charge of the electron and V is the potential difference.
The charge of an electron (e) is equal to -1.60 × 10^-19 C.
When the electron comes to rest, all of its kinetic energy is converted into potential energy. Therefore, the kinetic energy is equal to the potential energy.
Setting KE = PE, we can equate the formulas:
1/2 * m * v² = q * V
Plugging in the values, we have:
1/2 * (9.11 × 10^-31 kg) * (4.00 × 10^6 m/s)² = (-1.60 × 10^-19 C) * V
Now we can solve for V:
V = (1/2 * (9.11 × 10^-31 kg) * (4.00 × 10^6 m/s)²) / (-1.60 × 10^-19 C)
Calculating this expression, we find that the potential difference required to bring the electron to rest is approximately equal to -7.23 × 10^2 volts. The negative sign indicates that the potential difference must be negative, meaning it requires a decrease in electric potential energy to bring the electron to rest.
So, the answer to the question is a potential difference of -7.23 × 10^2 volts.