devarshi-dt-logo

Question:

A proton is fired from very far away towards a nucleus with charge Q = 120 e, where e is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is :(take the proton mass, mp=(5/3)×10⁻²⁷ kg and h/e=4.2×10⁻¹⁵ J.s/C; 1/4πε₀=9×10⁹ m/F; 1fm=10⁻¹⁵m)

8 fm

7 fm

10 fm

9 fm

Solution:

Given here, q₁=120×e q₂=e ( charge on proton)
It is based on conservation of energy of proton.
(1/2)mv² = kq₁q₂/R
v = √(2Kq₁q₂/mR)
Also, we know, de Broglie wavelength, λ=h/mv
λ=h√(R/2mKq₁q₂)
Thus,
λ = h√(10×10⁻¹⁵)/√(2×(5/3)×10⁻²⁷×9×10⁹×120×e²)
λ = h/e × 10⁻¹⁴ × 10⁻⁶
λ = 4.2×10⁻¹⁵ × 10⁻¹⁴ × 10⁶
λ = 7×10⁻¹⁵ m = 7 fm
Option A is correct.