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 :

Solution

Physics sir · 4 steps

The parent was at rest, so the two pieces fly apart with equal and opposite momenta, and λ = h/p.

1. Momentum is conserved. Before the break-up the particle is at rest, so the total momentum is zero. Afterwards it must still be zero, so the two pieces move in opposite directions.

   0 = M v₁ + 3M v₂ ⇒ |p₁| = |p₂| = p

2. Use de Broglie. The de Broglie wavelength is λ = h/p. It depends only on the size of the momentum, not on the mass or the speed separately.
3. Compare. Both pieces have the same p, so λ_M/λ_(3M) = (h/p)/(h/p) = 1. The lighter piece moves 3 times faster, but its momentum is the same.
4. So. The ratio of the wavelengths is 1 : 1, which is option (d).

The idea to keep. Whenever something at rest splits into two parts, the parts carry equal momenta. Since λ = h/p, their de Broglie wavelengths are always equal, whatever their masses.

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