A particle A of mass m and initial velocity v collides with a particle B of mass (m)/(2) which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λ _ A to λ _ B after the

Correct answer: C. λ _ A λ _ B = 2 mV = m V _ A + (m)/(2) V _ B V = V _ A + V _ B 2 - - - - - - - - - - - - (i) V = V _ B - V _ A - - - - - - - - - - - - (ii) Solving (i) and (ii) 2V = 3 V _ B 2 ∴ V _ B = (4V)/(3) V _ A = V _ B - V = (V)/(3) P _ A = (mV)/(3) P _ B = (m)/(2) . (4V)/(3) λ _ A λ _ B = h × P _ B P _ A × h = ((2mV)/(3))/((mV)/(3)) = 2

JEE Main 20172 April

Exam: JEE Main
Year: 2017
Date: 2 April
Subject: Physics
Chapter: Dual Nature of Matter

JEE Main 2017 · 2 April · Physics · Dual Nature of Matter

A particle A of mass m and initial velocity v collides with a...

A particle

A

of mass

m

and initial velocity

v

collides with a particle

B

of mass

\dfrac{m}{2}

which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths

{\lambda}_{A}

{\lambda}_{B}

after the collision is: A particle A of mass m and initial velocity v collides with a particle B of mass (m)/(2) which is at rest. The

Options

A $\dfrac{{\lambda}_{A}}{{\lambda}_{B}} = \dfrac{1}{2}$
B $\dfrac{{\lambda}_{A}}{{\lambda}_{B}} = \dfrac{1}{3}$
C $\dfrac{{\lambda}_{A}}{{\lambda}_{B}} = 2$
D $\dfrac{{\lambda}_{A}}{{\lambda}_{B}} = \dfrac{2}{3}$

Correct answer

C. $\dfrac{{\lambda}_{A}}{{\lambda}_{B}} = 2$

Step-by-step solution

mV = m{V}_{A} + \dfrac{m}{2} {V}_{B}

V = {V}_{A} + \dfrac{{V}_{B}}{2} - - - - - - - - - - - - \text{(i)}

V = {V}_{B} - {V}_{A} - - - - - - - - - - - - \text{(ii)}

\text{Solving (i) and (ii)}

2V = \dfrac{3{V}_{B}}{2}

∴ {V}_{B} = \dfrac{4V}{3}

{V}_{A} = {V}_{B} - V = \dfrac{V}{3}

{P}_{A} = \dfrac{mV}{3}

{P}_{B} = \dfrac{m}{2} . \dfrac{4V}{3}

\dfrac{{λ}_{A}}{{λ}_{B}} = \dfrac{h \times {P}_{B}}{{P}_{A} \times h} = \dfrac{\dfrac{2mV}{3}}{\dfrac{mV}{3}} = 2

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