Question

Modern Physics: Experimental basis for quantum theory + Atomic models

The actual size of a gold nucleus is approximately 7 fm. Knowing this, calculate the kinetic energy (KE, or KE_a, however you may choose to label the measurement) that an alpha particle would need to just touch the outside of the nucleus. Does this seem like a reasonable number?

Hint - the LHC is currently the most powerful particle accelerator on Earth, and i?t operates at 13 TeV (TeV = 10¹² eV).

Answer 1

Acc. To formula for distance of closest approach

Kinetic energy of alpha particle = Potential energy between alpha and gold atom

K.E. = (1/ 4×3.14€₀) q Q / d

Q= 79 ×1.6×10^-19 C

q= 2× 1.6 ×10^-19 C

d= 7 fermi = 7×10^-15 m

(1/ 4×3.14€₀ )= 9 ×10⁹

K.E.= (9 × 10⁹ ×79 ×2 ×1.6× 1.6 ×10^-38 )/ (7×10^-15 )

= 520.04571 × 10^-14

Or K.E. = 520.04571× 10 ^-14 / 1.6 × 10^-19

K.E. = 325.02857 × 10⁵ eV

K.E. = 32117 TeV

which is very high as compare to the operating energy of LHC so this is not a reasonable number .