A proton is fired from very far away towards a nucleus with charge Q = 120 e , where e is the electronic charge. It makes the closest approach of 10 fm to the nucleus. The de Broglie wavelength (in fm ) of the proton at its start is: ( m p = 5 / 3 × 1 0 - 2 7 kg ; h e = 4 · 2 × 1 0 - 1 5 J s C - 1 ; 1 4 π ε 0 = 9 × 1 0 9 N m 2 C - 2 ; 1 fm = 1 0 - 1 5 m)

Step-by-step solution

0 + 1 2 mv 2 = K Q e 1 0 × 1 0 - 1 5 = K 1 2 0 e e 1 0 × 1 0 - 1 5 1 2 × 5 3 × 1 0 - 2 7 v 2 = 9 × 1 0 9 × 1 2 0 × 1 · 6 × 1 0 - 1 9 2 1 0 × 1 0 - 1 5 v = 9 × 6 × 1 0 9 × 1 2 0 × 2 · 5 6 × 1 0 - 3 8 5 0 × 1 0 - 4 2 v = 3 3 1 · 7 7 6 × 1 0 1 3 λ = h mv λ = 4 · 2 × 1 0 - 1 5 × 1 · 6 × 1 0 - 1 9 5 3 × 1 0 - 2 7 × 3 3 1 · 7 7 6 × 1 0 1 3 = 4 · 2 × 4 · 8 × 1 0 - 3 4 5 7 · 6 × 5 × 1 0 - 2 1 = 0 · 0 7 × 1 0 - 1 3 λ = 7 × 1 0 - 1 5 = 7 fm