Particle A of mass m _ A = (m)/(2) moving along the x -axis with velocity v _ 0 collides elastically with another particle B at rest having mass m _ B = (m)/(3) . If both the particles move along the x -axis after the

Correct answer: D. ∆λ = 4 λ _ 0 v _ 1 = (2(m / 3)0)/(((m)/(2) + (m)/(3))) + (((m)/(2) - (m)/(3))v)/(((m)/(2) + (m)/(3))) = (v)/(5) For particle A , Initial de-Broglie wavelength λ _ 0 = (h)/((m)/(2)v) = (2h)/(mv) Final de-Broglie wavelength after collision. λ _ 1 = (h)/((m)/(2)(v)/(5)) = (10h)/(mv) = 5 λ _ 0 Change in De-Broglie wavelength Δ λ = λ _ i - λ _ 0 = 4 λ _ 0

JEE Main 20204 SeptemberMorning Shift

Exam: JEE Main
Year: 2020
Date: 4 September
Shift: Morning Shift
Subject: Physics
Chapter: Dual Nature of Matter

JEE Main 2020 · 4 September · Morning Shift · Physics · Dual Nature of Matter

Particle A of mass m _ A = (m)/(2) moving along the x -axis with...

Particle

A

of mass

{m}_{A} = \dfrac{m}{2}

moving along the

\text{x}

-axis with velocity

{v}_{0}

collides elastically with another particle

\text{B}

at rest having mass

{m}_{B} = \dfrac{m}{3}

. If both the particles move along the

\text{x}

-axis after the collision, the change

∆ \lambda

in the wavelength of the particle

\text{A}

, in terms of its de-Broglie wavelength

\left({λ}_{0}\right)

before the collision is: Particle A of mass m _ A = (m)/(2) moving along the x -axis with velocity v _ 0 collides elastically with another

Options

A $∆\lambda = \dfrac{3}{2}{\lambda}_{0}$
B $∆\lambda = \dfrac{5}{2}{\lambda}_{0}$
C $∆\lambda = 2{\lambda}_{0}$
D $∆\lambda = 4{\lambda}_{0}$

Correct answer

D. $∆\lambda = 4{\lambda}_{0}$

Step-by-step solution

Particle A of mass m _ A = (m)/(2) moving along the x -axis with velocity v _ 0 collides elastically with another

${v}_{1} = \dfrac{2\left(m / 3\right)0}{\left(\dfrac{m}{2} + \dfrac{m}{3}\right)} + \dfrac{\left(\dfrac{m}{2} - \dfrac{m}{3}\right)v}{\left(\dfrac{m}{2} + \dfrac{m}{3}\right)} = \dfrac{v}{5}$

For particle $A$ ,

Initial de-Broglie wavelength

${\lambda}_{0} = \dfrac{h}{\dfrac{m}{2}v} = \dfrac{2h}{mv}$

Final de-Broglie wavelength after collision.

${\lambda}_{1} = \dfrac{h}{\dfrac{m}{2}\dfrac{v}{5}} = \dfrac{10h}{mv} = 5{\lambda}_{0}$

Change in De-Broglie wavelength $\Delta \lambda = {\lambda}_{i} - {\lambda}_{0} = 4{\lambda}_{0}$

Practice Dual Nature of Matter on Quantrex Academy →

More from Dual Nature of Matter

K_1 and K_2 be the maximum kinetic energies of photoelectrons emitted from a surface of a… JEE Main 2026 The de Broglie wavelength for an electron accelerated through the potential difference of… JEE Main 2026 Given below are two statements: one is labelled as Assertion (A) and the other is… JEE Main 2026 An electron is travelling with a velocity v in free space and when it enters a medium… JEE Main 2026 An electron of mass m is moving in an electric field E = -2E_0 i ( E_0 = constant > 0 )… JEE Main 2026 Light source having wavelength 331 nm is used to generate photo-electrons whose stopping… JEE Main 2026 Match the LIST-I with LIST-II List-I List-II A. Planck's constant I. ML^2T^ -2 B.… JEE Main 2026 The de Broglie wavelength associated with an electron accelerated through a potential… JEE Main 2026 Full Dual Nature of Matter list All JEE Main PYQs