Question

In one of the classic nuclear physics experiments performed by Ernest Rutherford at the beginning of the 20th century, alpha particles (helium nuclei) were shot at gold nuclei and their paths were substantially affected by the Coulomb repulsion from the nuclei. If the energy of the (doubly charged) alpha nucleus was 5.1 MeV, how close to the gold nucleus (79 protons) could it come before being deflected? r =

Answer 1

Answer:

r  = 3.8 × 10 ⁻¹⁴ m

Explanation:

given data

alpha nucleus = 5.1 MeV

Charge of the alpha particle q₁= 2 × 1.6 × 10⁻¹⁹ C = 3.2 × 10⁻¹⁹ C

Charge of the gold nucleus q₂= 79 × 1.6 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷ C

Kinetic energy of  the alpha particle = 5.97 × 10⁶ × 1.602 × 10⁻¹⁹ J ( 1 eV) =  9.564 × 10⁻¹³

k electrostatic force constant = 9 × 10⁹ N.m²/c²

solution

we know that when its kinetic energy is equal to the potential energy than  alpha particle will deflect \

so

Kinetic energy = potential energy =   k q₁q₂ ÷ r   ..................1

here  r is close distance the alpha particle

so r will be put here value

r = (  9 × 10⁹  × 3.2 × 10⁻¹⁹  × 1.264 × 10⁻¹⁷ ) ÷ ( 9.564 × 10⁻¹³  )

r  = 3.8 × 10 ⁻¹⁴ m