The potential difference between two parallel plates is 458 V. An alpha particle, having a mass of 6.64 × 10^−27 kg and a charge of 3.20 × 10^−19 C, is released from rest near the positive plate. What is the speed of the alpha particle when it reaches the other plate, given that the distance between the plates is 40.6 cm?
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To calculate the speed of the alpha particle when it reaches the other plate, we can use the concept of the conservation of energy. When the alpha particle is released from rest, its initial kinetic energy (K.E.) is zero, and its entire energy is in the form of electric potential energy (P.E.).
The electric potential energy (P.E.) can be calculated as: \[ P.E. = qV \] Where: - \( q \) is the charge of the particle - \( V \) is the potential difference between the plates
The kinetic energy (K.E.) of a particle with mass \( m \) and velocity \( v \) is given by: \[ K.E. = \frac{1}{2} mv^2 \]
As the particle accelerates from the positive plate to the negative plate, its potential energy is converted into kinetic energy. The final kinetic energy at the negative plate will be equal to the initial potential energy at the positive plate, assuming no energy losses.
Therefore, we can set the initial P.E. equal to the final K.E.: \[ qV = \frac{1}{2} mv^2 \]
We can now solve for \( v \): \[ v = \sqrt{\frac{2qV}{m}} \]
Now we will plug in the values: \[ q = 3.20 \times 10^{-19} \, \text{C} \] \[ V = 458 \, \text{V} \] \[ m = 6.64 \times 10^{-27} \, \text{kg} \]
\[ v = \sqrt{\frac{2 \times 3.20 \times 10^{-19} \, \text{C} \times 458 \, \text{V}}{6.64 \times 10^{-27} \, \text{kg}}} \]
\[ v = \sqrt{\frac{2 \times 3.20 \times 10^{-19} \times 458}{6.64 \times 10^{-27}}} \]
\[ v = \sqrt{\frac{2 \times 3.20 \times 458 \times 10^{-19}}{6.64 \times 10^{-27}}} \]
\[ v = \sqrt{\frac{2924.8 \times 10^{-19}}{6.64 \times 10^{-27}}} \]
\[ v \approx \sqrt{4.40458 \times 10^8} \]
\[ v \approx 20963.9 \, \text{m/s} \]
So the speed of the alpha particle when it reaches the other plate is approximately \( 20963.9 \, \text{m/s} \).