A particle of mass 5M at rest decays into two particle of mass 2m and 3m having non zero velocities the ratio of debroglie wavelengths of particle is

a] 3/2 b] 2/3 c] 1/3 d] none of these

I got the ans 3/2 but the book says the ans is d

can anyone give me a detailed ans and concept behind it if any

I promise to rate everyone who helps me

To tackle the problem of a particle of mass 5M decaying into two particles of masses 2m and 3m, we need to apply the principles of conservation of momentum and energy, along with the concept of de Broglie wavelength. Let's break this down step by step to clarify the situation and find the correct ratio of the de Broglie wavelengths of the two resulting particles.

Understanding the Decay Process

Initially, we have a particle at rest with a total mass of 5M. When it decays into two particles, one with mass 2m and the other with mass 3m, we can denote their velocities as v₁ for the 2m particle and v₂ for the 3m particle.

Applying Conservation of Momentum

Since the initial particle is at rest, the total momentum before the decay is zero. According to the conservation of momentum, the total momentum after the decay must also equal zero:

• Momentum of 2m particle: 2m * v₁
• Momentum of 3m particle: 3m * v₂

Setting the total momentum to zero gives us:

2m * v₁ + 3m * v₂ = 0

This can be rearranged to:

2v₁ + 3v₂ = 0

From this equation, we can express v₁ in terms of v₂:

v₁ = - (3/2) v₂

Using Conservation of Energy

The total energy before the decay is simply the rest energy of the original particle, given by:

E_initial = 5Mc²

After the decay, the total energy is the sum of the kinetic energies of the two particles plus their rest energies:

E_final = 2mc² + 3mc² + (1/2)(2m)v₁² + (1/2)(3m)v₂²

Setting the initial energy equal to the final energy gives us:

5Mc² = 5mc² + (1/2)(2m)v₁² + (1/2)(3m)v₂²

Calculating the De Broglie Wavelengths

The de Broglie wavelength (λ) of a particle is given by the formula:

λ = h / p

where h is Planck's constant and p is the momentum of the particle. For each particle, we can express the momentum as:

• p₁ = 2m * v₁
• p₂ = 3m * v₂

Thus, the de Broglie wavelengths become:

λ₁ = h / (2m * v₁)

λ₂ = h / (3m * v₂)

Finding the Ratio of Wavelengths

The ratio of the de Broglie wavelengths is:

λ₁ / λ₂ = (h / (2m * v₁)) / (h / (3m * v₂))

This simplifies to:

λ₁ / λ₂ = (3v₂) / (2v₁)

Substituting v₁ = - (3/2) v₂ into the ratio gives:

λ₁ / λ₂ = (3v₂) / (2 * (-3/2)v₂) = (3v₂₂) = -1

Since we are interested in the absolute value of the ratio, we find:

λ₁ / λ₂ = 1

Final Thoughts

In conclusion, the ratio of the de Broglie wavelengths of the two particles is indeed not 3/2, but rather 1. This aligns with the conservation laws and the calculations we've performed. Therefore, the answer provided in your book, which states that the answer is "none of these," is correct. If you have any further questions or need clarification on any part of this explanation, feel free to ask!