A proton and a Li³⁺ nucleus are accelerated by the same potential. If λ _(Li) and λ ₚ denote the de Broglie wavelengths of Li³⁺ and proton respectively, then the value of (λ _(Li))/(λ ₚ) is x × 10⁻¹.
The value of x is ______. (Rounded off to the nearest integer)
[Mass of Li³⁺ = 8.3 mass of proton]

Solution

Chemistry ma’am · 3 steps

Same voltage does not mean same energy: the ion with charge 3 picks up three times the kinetic energy.

1. Energy from voltage. A charge q accelerated through potential V gains kinetic energy qV. Its momentum is p = √2mqV, so λ = h/p depends on both mass and charge.

   λ = h/√2mqV ⇒ λ_(Li)/λₚ = √((mₚ qₚ)/(m_(Li) q_(Li)))

2. Put in numbers. h and V cancel. With m_(Li) = 8.3 mₚ and q_(Li) = 3e, qₚ = e, the ratio is √(1/(8.3 × 3)) = √(1/24.9) = 1/4.99 = 0.2004.
3. So. λ_(Li)/λₚ ≈ 2.0 × 10⁻¹, so x rounds to 2.

The common slip: 3. Leaving out the charge of Li³⁺ gives √(1/8.3) = 0.347, so x comes out as 3, not 2.

The idea to keep. For particles accelerated through a potential, use λ = h/√2mqV. Always check the charge: a heavier, more charged ion has a much shorter wavelength.

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