JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Modern Physics

A proton and an

α

-particle are accelerated from rest by 2 V and 4 V potentials, respectively. The ratio of their de-Broglie wavelength is:

(A) $4 : 1$ ✓ Correct
(B) $2 : 1$
(C) $8 : 1$
(D) $16 : 1$

Answer: (A)

Solution

The de-Broglie wavelength is

λ = \frac{h}{2 m q V}

. The ratio of wavelengths is

\frac{λ _{p}}{λ _{α}} = \frac{m _{α} q _{α} V _{α}}{m _{p} q _{p} V _{p}}

. For a proton:

m_{p} = m

q_{p} = e

V_{p} = 2

V. For an

α

-particle:

m_{α} = 4 m

q_{α} = 2 e

V_{α} = 4

V. Thus,

\frac{λ _{p}}{λ _{α}} = \frac{4 m \cdot 2 e \cdot 4}{m \cdot e \cdot 2} = \frac{32}{2} = 16 = 4

. Therefore,

λ_{p} : λ_{α} = 4 : 1

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Single correct option

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