JEE Advanced2012PhysicsDual Nature of MatterActual
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, m_ p =(5 / 3) 10⁻²⁷ ~kg ; h / e=4.2 10⁻¹⁵ ~J . S / C ; 1 4 ₀ =9 10⁹ ~m / F ; 1 fm =10⁻¹⁵ ~m
Correct answer
7
Step-by-step solution
From energy conservation Loss in K.E. of proton = gain in potential energy of the proton - nucleus system array c 1 2 m v²= 1 4 ₀ q₁ q₂ r p² 2 m = 1 4 ₀ q₁ q₂ r 1 2 m ( h² ² )= 1 4 ₀ q₁ q₂ r = 4 ₀ r h² q₁ q₂(2 m) array Putting the values of 4 ₀, r, h, q₁, q₂ and m we get, deBroglie wavelength of proton, =7 fm