The kinetic energy of an electron, α-particle and a proton are given as 4 K, 2 K and K respectively. The de-Broglie wavelength associated with electron (λ e), α-particle ((λ α) and the proton (λ p) are as follows:

Solution

Physics sir · 3 steps

With kinetic energy given, λ = h/√2mK, so just compare mass times energy for each particle.

1. Link λ to energy. Momentum and kinetic energy are related by p = √2mK. Put this into λ = h/p.

   λ = h/√2mK ⇒ λ ∝ 1/√mK

2. Compare mK. Take the proton mass as m. Then m_α = 4m and mₑ ≈ m/1836. A bigger product mK means a smaller wavelength.

   Particle	Mass	Kinetic energy	Product mK	Wavelength
   α-particle	4m	2K	8mK, largest	smallest
   Proton	m	K	mK	middle
   Electron	m/1836	4K	mK/459, smallest	largest

3. So. λ_α = λₚ/√8 ≈ 0.35 λₚ and λₑ = √459 λₚ ≈ 21 λₚ. The order is λ_α < λₚ < λₑ, which is option (a).

The idea to keep. When kinetic energy is given, use λ = h/√2mK. Heavier and more energetic means shorter wavelength, and the electron's tiny mass beats any small energy advantage.

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