If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is m. (Atomic number of gold =79 and 1/(4πε₀)=9×10⁹ in SI units)

Question

Accepted Answer

(2) 2.95×10⁻¹⁴. At closest approach all KE becomes PE: K=1/(4πε₀)((2e)(79e))/(r₀). K=7.7×1.6×10⁻¹³=1.232×10⁻¹² J. r₀=(9×10⁹×158×(1.6×10⁻¹⁹)²)/(1.232×10⁻¹²)=2.95×10⁻¹⁴ m.

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ______ m. (Atomic number of gold =79 and 1/(4πε₀)=9×10⁹ in SI units)

Step-by-step solution

Why the other options are wrong