An electron accelerated through a potential difference V₁ has a de-Broglie wavelength of λ. When the potential is changed to V₂, its de-Broglie wavelength increases by 50 %. The value of (V₁/V₂) is equal to

Solution

Physics sir · 3 steps

Wavelength goes as one over root V, so square the wavelength ratio to get the voltage ratio.

1. Wavelength and voltage. An electron accelerated through V gains kinetic energy eV, so its momentum is p = √2meV. Then λ = h/p = h/√2meV, which means λ ∝ 1/√V.

   λ₂/λ₁ = √(V₁/V₂) ⇒ V₁/V₂ = (λ₂/λ₁)²

2. Use the 50% increase. A 50% increase means λ₂ = 1.5λ = 3/2λ, so λ₂/λ₁ = 3/2.
3. So. V₁/V₂ = (3/2)² = 9/4, which is option (d). A longer wavelength needs a smaller voltage, so V₁ > V₂ fits.

The idea to keep. For a charged particle accelerated from rest, λ = h/√2mqV. Any ratio of voltages is the square of the inverse ratio of wavelengths.

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