NTA Abhyas JEE Main2020ChemistryStructure of AtomPractice

Calculate the ratio of wavelength for an α particle and proton accelerated through same potential difference.

Options

A1 2
B1 2
C1 2 2
D2 2

Correct answer

C. 1 2 2

Step-by-step solution

Using de Broglie equation λ = h 2 q m V λ = wavelength, q = charge, m = mass, V = Potential and h = Plank constant λ a = h 2 q a m a V λ p = h 2 q p m p V m a = 4 m p , q a = 2 q p λ a λ p = q p m p q a m a = 1 2 . 1 4 = 1 8 λ a λ p = 1 2 2

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