Question

The potential difference between two parallel plates is 458 V. An alpha particle with mass of 6.64 ×10−27 kg and 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? The distance between the plates is 40.6 cm.

Answer 1

21.01 × 10⁴ m/s

Step-by-step explanation:

Data provided in the question:

Potential difference between two parallel plates, V = 458 V

Mass of the alpha particle = 6.64 × 10⁻²⁷ kg

Charge released, q = 3.20 × 10⁻¹⁹ C

Distance between the plates, d = 40.6 cm

Now,

qV =
`(1)/(2)mv^2`

on substituting the respective values, we get

(3.20 × 10⁻¹⁹) × 458 =
`(1)/(2)*(6.64*10^(-27))v^2`

or

1465.6 × 10⁻¹⁹ = (3.32 × 10⁻²⁷)v²

or

v = 21.01 × 10⁴ m/s