Electron beam used in an electron microscope, when accelerated by a voltage of 20 kV, has a de-Broglie wavelength of λ₀. IF the voltage is increased to 40 kV, then the de-Broglie wavelength associated with the electron beam would be :

Solution

Physics sir · 3 steps

The wavelength goes as one over root V, so doubling the voltage shrinks it by √2, not by 2.

1. Momentum from voltage. An electron accelerated through a potential difference V gains kinetic energy eV. Since K = p²/2m, its momentum is p = √2meV.

   λ = h/p = h/√2meV ⇒ λ ∝ 1/√V

2. Take the ratio. Here h, m and e are the same for both beams, so they cancel. That gives λ/λ₀ = √((20 kV)/(40 kV)) = 1/√2.
3. So. The new wavelength is λ₀/√2, which is option (c).

The idea to keep. For a charge accelerated through V from rest, λ ∝ 1/√V. Always check whether λ goes with V or with √V before you scale it.

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