Question

A particle of mass 4M at rest disintegrates into two particles of mass M and 3M respectively having non zero velocities. The ratio of de-Broglie wavelength of particle of mass M to that of mass 3M will be :

• A. 1 : 1
• B. 1 : 3
• C. 3 : 1
• D. 1 : $\sqrt{3}$

Hint:

In any internal explosion or disintegration from rest, the fragments always have equal and opposite momenta.
Consequently, their de-Broglie wavelengths are always identical.

Answer 1

The Correct Option is A

Given:

A particle of mass 4M at rest disintegrates into two particles of masses M and 3M.

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Step 1: Apply conservation of momentum

Initial momentum = 0 (since particle is at rest)

Let velocity of particle of mass M = v₁
Let velocity of particle of mass 3M = v₂

Using conservation of momentum:

Mv₁ + 3Mv₂ = 0

v₁ = −3v₂

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Step 2: Calculate momenta of both particles

Momentum of particle of mass M:

p₁ = Mv₁ = M(−3v₂) = −3Mv₂

Momentum of particle of mass 3M:

p₂ = 3Mv₂

Hence,

|p₁| = |p₂| = 3Mv₂

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Step 3: Use de-Broglie relation

λ = h / p

λ₁ = h / |p₁| = h / (3Mv₂)

λ₂ = h / |p₂| = h / (3Mv₂)

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Step 4: Find ratio

λ₁ / λ₂ = 1

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Final Answer:

The ratio of de-Broglie wavelengths is,
1 : 1