A bare helium nucleus has two positive charges and a mass of 6.64 ✕ 10−27 kg. (a) Calculate its kinetic energy in joules at 3.60% of the speed of light.

Respuesta :

Answer:

the kinetic energy of the particle is 3.872 x 10⁻¹³ J

Explanation:

Given;

mass of helium, m = 6.64 x 10⁻²⁷ kg

speed of light, c = 3 x 10⁸ m/s

3.6% of speed of light, c = 0.036 x 3 x 10⁸ m/s = 1.08 x 10⁷ m/s

The kinetic energy of the particle is calculated as;

K.E = ¹/₂mc²

K.E = ¹/₂ x (6.64 x 10⁻²⁷) x (1.08 x 10⁷)²

K.E = 3.872 x 10⁻¹³ J

Therefore, the kinetic energy of the particle is 3.872 x 10⁻¹³ J