Verified Solution • 2025–26 Session moderate

QJEEM17Phfae4: JEE Main 2017 Physics - Class 12 JEEM17Phfae4 Solution: 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 m/2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λA{\lambda _A} to λB{\lambda _B} after the collision is:

Multiple Choice Options

  • A λAλB=13{{{\lambda _A}} \over {{\lambda _B}}} = {1 \over 3}
  • B λAλB=2{{{\lambda _A}} \over {{\lambda _B}}} = 2
  • C λAλB=23{{{\lambda _A}} \over {{\lambda _B}}} = {2 \over 3}
  • D λAλB=12{{{\lambda _A}} \over {{\lambda _B}}} = {1 \over 2}

Step-by-Step Verified Solution

From question, mA = M; mB = m2{m \over 2}

uA = V and uB = 0

Let after collision velocity of A = V1 and

velocity of B = V2

Applying law of conservation of momentum,

mu = mv1 + (m2)v2\left( {{m \over 2}} \right){v_2}

⇒\Rightarrow 24= 2v1 + v2 ........(i)

By law of collision

e=v2−v1u−0e = {{{v_2} - {v_1}} \over {u - 0}}

⇒\Rightarrow u = v2 - v1 ..........(ii)

[As collision is elastic, e = 1]

using eqns (i) and (ii)

v1 = 43{4 \over 3} and v2 = 43v{4 \over 3}v

We know, de-Broglie wavelength λ\lambda = hp{h \over p}

∴\therefore λAλB=PBPA{{{\lambda _A}} \over {{\lambda _B}}} = {{{P_B}} \over {{P_A}}} = m2×43um×43{{{m \over 2} \times {4 \over 3}u} \over {m \times {4 \over 3}}} = 2

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