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