JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Modern Physics

Particle A of mass

m_{A} = \frac{m}{2}

moving along the x-axis with velocity

v_{0}

collides elastically with another particle

B

at rest having mass

m_{B} = \frac{m}{3}

. If both particles move along the

x

axis after the collision, the change

Δ λ

in de-Broglie wavelength of particle A, in terms of its de-Broglie wavelength (

λ_{0}

) before collision is:

Answer: (D)

Solution

For a one-dimensional elastic collision the velocity of

A

afterwards is

v_{1} = 2 u_{c m} - v_{0}

, where the centre-of-mass velocity is

u_{c m} = \frac{m _{A} v _{0}}{m _{A} + m _{B}} = \frac{( m /2 ) v _{0}}{m /2 + m /3} = \frac{3 v _{0}}{5}

. Hence

v_{1} = 2 \cdot \frac{3 v _{0}}{5} - v_{0} = \frac{v _{0}}{5}

. The de Broglie wavelength is inversely proportional to speed, so with the speed cut to one fifth the wavelength grows fivefold, from

λ_{0} = \frac{h}{m _{A} v _{0}}

λ_{f} = \frac{h}{m _{A} v _{1}} = 5 λ_{0}

. The change is

Δ λ = λ_{f} - λ_{0} = 5 λ_{0} - λ_{0} = 4 λ_{0}

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option