Question

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 de-broglie wavelengths λa and λb after the collision is

Answer 1

The ratio of their de-broglie wavelength will be 2:1

Explanation:

Let after collision the velocity be v₁ and v₂

From the laws of conservation of momentum

mv = mv₁ + mv₂/2

=> 2v = 2v₁ + v₂.................eq1

Also since the collision is elastic we have,

1 = (v₂-v₁)/(v-0)

=> v = v₂ - v₁........eq2

solving both the equation we get

v₁ = v/3 and v₂ = 4v/3

we know that the de-broglie wavelength is given by,

λ = h/p

where p = momentum = mv

h = plank's constant

For particle a

λa = h/mv₁ = 3h/mv

for particle b

λb = h/(m/2)(4v/3) = 3h/2mv

Hence the ratio of their de-broglie wavelength

= λa:λb = 2:1