An α particle and a carbon 12 atom has same kinetic energy K. The ratio of their de-Broglie wavelengths (λ _α :λ _(C12)) is :

Solution

Physics sir · 3 steps

Same kinetic energy means λ goes as 1/√m, so the lighter alpha gets the longer wavelength.

1. Link λ to K. Kinetic energy and momentum are related by K = p²/2m, so p = √2mK. Put this into λ = h/p.

   λ = h/√2mK ⇒ λ_α/λ_C = √(m_C/m_α)

2. Put in masses. With the same K for both, h and K cancel. The alpha particle has mass 4 u and the carbon-12 atom has mass 12 u, so the ratio is √(12/4) = √3.
3. So. λ_α : λ_(C12) = √3 : 1, which is option (b).

The idea to keep. For equal kinetic energy, λ = h/√2mK gives λ ∝ 1/√m. The heavier particle always has the shorter wavelength. Mass sits in the denominator and under a square root.

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