Two particles move at right angle to each other. Their de Broglie wavelengths are $\lambda_{1}$ and $\lambda_{2}$ respectively. The particles suffere perfectly inelastic collision. The de Broglie wavelength $\lambda$, of the final particle, is given by:

(D) $\frac{1}{\lambda^{2}}=\frac{1}{\lambda_{1}^{2}}+\frac{1}{\lambda_{2}^{2}}$

(A) $\lambda=\sqrt{\lambda_{1} \lambda_{2}}$

(B) $\lambda=\frac{\lambda_{1}+\lambda_{2}}{2}$

(C) $\frac{2}{\lambda}=\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}$

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Modern Physics

Two particles move at right angle to each other. Their de Broglie wavelengths are

λ_{1}

and

λ_{2}

respectively. The particles suffere perfectly inelastic collision. The de Broglie wavelength

λ

, of the final particle, is given by:

(A) $λ = λ_{1} λ_{2}$
(B) $λ = \frac{λ _{1} + λ _{2}}{2}$
(C) $\frac{2}{λ} = \frac{1}{λ _{1}} + \frac{1}{λ _{2}}$
(D) $\frac{1}{λ ^{2}} = \frac{1}{λ _{1}^{2}} + \frac{1}{λ _{2}^{2}}$ ✓ Correct

Answer: (D)

Solution

By de Broglie each particle carries momentum

p = \frac{h}{λ}

, so before the collision the two momenta are

\frac{h}{λ _{1}}

and

\frac{h}{λ _{2}}

directed at right angles. In a perfectly inelastic collision the particles stick and momentum is conserved as a vector, so the combined momentum has magnitude

p = (\frac{h}{λ _{1}})^{2} + (\frac{h}{λ _{2}})^{2}

. Writing

p = \frac{h}{λ}

for the final particle and squaring gives

\frac{1}{λ ^{2}} = \frac{1}{λ _{1}^{2}} + \frac{1}{λ _{2}^{2}}

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Modern Physics

Solution

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