Answer:
21.01 × 10⁴ m/s
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 = [tex]\frac{1}{2}mv^2[/tex]
on substituting the respective values, we get
(3.20 × 10⁻¹⁹) × 458 = [tex]\frac{1}{2}\times(6.64\times10^{-27})v^2[/tex]
or
1465.6 × 10⁻¹⁹ = (3.32 × 10⁻²⁷)v²
or
v = 21.01 × 10⁴ m/s
