contestada

In a head-on collision, an alpha particle (Z = 2) of energy 8.80 MeV bounces straight back from a nucleus of charge 82.0 e. How close were the centers of the objects at closest approach?

Respuesta :

Answer:

2.7 x 10^-14 m

Explanation:

E =8.8 MeV = 8.8 x 1.6 x 10^-13 J

q = 2 e = 2 x 1.6 x 10^-19 C

Q = 82 e = 82 x 1.6 x 10^-19 C

Let d be the distance of closest approach

E = k Q q / d

Where, K = 9 x 10^9 Nm^2 / C^2

d = k Q q / E

d = (9 x 10^9 x 82 x 1.6 x 10^-19 x 2 x 1.6 x 10^-19) / (8.8 x 1.6 x 0^-13)

d = 2.7 x 10^-14 m

Otras preguntas

Q&A Education