contestada

In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated towards a gold nucleus and its path was substantially deflected by the Coulomb interaction. If the initial kinetic energy of the doubly charged alpha nucleus was 5.21 MeV, how close to the gold nucleus (79 protons) could it come before being turned around?

Respuesta :

Answer:

here distance is 4.45 [tex]10^{-14}[/tex] meter

Explanation:

Given data

alpha particle accelerated towards a gold nucleus

alpha nucleus = 5.21 MeV

to find out

how close to the gold nucleus

solution

we know according to conversation of energy that

initial kinetic energy is equal to final potential energy ( electrostatic)  i.e

1/2 m v² = k  q(a) q(b) / d

here we find d by put all value q(a) q(b) and k amd m v

we get

d = 2 × 9 × [tex]10^{9}[/tex] × 2× 79 × (1.6 ×[tex]10^{-19})^2[/tex]  / 2 × 5.1  × [tex]10^{6}[/tex] ×1.6× [tex]10^{-19}[/tex]

so here distance is 4.45 [tex]10^{-14}[/tex] meter

Otras preguntas