The ratio of wavelengths of proton and deuteron accelerated by potential V p and V d is 1 : 2 . Then, the ratio of V p to V d will be

Step-by-step solution

Kinetic energy gained by a charged particle accelerated by a potential V is q V K E = q V Using relation between momentum and kinetic energy ⇒ p 2 2   m = q V ⇒ p = 2 mqV We know that momentum of charged particle is p = h λ . Thus, λ = h 2 m q V Now, λ p λ d = m d V d m p V p ⇒ 1 2 = 2 V d V p ⇒ V p V d = 4