Question

Two particles move at a right angle to each other. Their de-Broglie wavelengths are λ₁ and λ₂ respectively. The particles suffer a perfectly inelastic collision. The de-Broglie wavelength λ, of the final particle, is given by ______.

पर्याय

• `1/lambda^2 = 1/lambda_1^2 + 1/lambda_2^2`

• `lambda = sqrt(lambda_1lambda_2)`

• `lambda = (lambda_1 + lambda_2)/2`

• `2/lambda = 1/lambda_1 + 1/lambda_2`

Answer 1

Two particles move at a right angle to each other. Their de-Broglie wavelengths are λ₁ and λ₂ respectively. The particles suffer a perfectly inelastic collision. The de-Broglie wavelength λ, of the final particle, is given by `underlinebb(1/lambda^2 = 1/lambda_1^2 + 1/lambda_2^2)`.

Explanation:

Given: Particle 1 has a de-Broglie wavelength of λ₁, particle 2 has a de-Broglie wavelength of λ₂, both particles are travelling at a right angle to one another, and there is a perfectly inelastic collision between the two particles.

To find: The final particle's de-Broglie wavelength λ.

Particle-1's momentum is: 

`p_1 = h/lambda_1`

Particle-2's momentum is: 

`p_2 = h/lambda_2`

Before the collision, the two-particle system's total momentum was as follows:

`p = sqrt(p_1^2 + p_2^2 + 2p_1p_2 cos90^circ)` ......(i)

The two particles cling together and continue to move as one after the collision because the collision is inelastic. The last particle after a collision will have the following momentum according to the law of conservation of linear momentum:

`p = sqrt(p_1^2 + p_2^2 + 2p_1p_2 cos90^circ)`

= `sqrt((h/lambda_1)^2 + (h/lambda_2)^2 + 2(h/lambda_1)(h/lambda_2)cos90^circ)`

`p = h/lambda = hsqrt((1/lambda_1)^2 + (1/lambda_2)^2)`

`1/lambda = sqrt((1/lambda_1)^2 + (1/lambda_2)^2)`

`1/lambda^2 = 1/lambda_1^2 + 1/lambda_2^2`