Particle A of mass m A = m/2 moving along the x-axis with velocity v₀ collides elastically with another particle B at rest having mass m B = m/3. If both particles move along the x-axis after the collision, the change Δλ in de-Broglie wavlength of particle A, in terms of its de-Broglie wavelength (λ₀) before collision is :

Solution

Physics sir · 3 steps

Find A's speed after the elastic collision; since λ = h/mv with the same mass, λ scales as 1/v.

1. Speed of A after. In a head-on elastic collision with B at rest, the velocity of A afterwards comes from the standard result below.

   v_A = (m_A - m_B)/(m_A + m_B) v₀ = (m/2 - m/3)/(m/2 + m/3) v₀ = (1/6)/(5/6) v₀ = v₀/5

2. New wavelength. λ = h/(m_A v), and the mass of A does not change. So λ is proportional to 1/v. The speed drops to v₀/5, so the wavelength becomes λ₁ = 5λ₀.
3. So. Δλ = λ₁ - λ₀ = 5λ₀ - λ₀ = 4λ₀, which is option (d).

The idea to keep. For the same particle, λ = h/mv means the wavelength changes inversely with speed. Get the speed ratio from the elastic-collision formula, then read off the change in wavelength.

Answer from the NTA final answer key. Solution by PrepWiser.