The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through same potential of 100 V. What should nearly be the ratio of their wavelengths? (mₚ = 1.00727u mₑ = 0.00055u)

Solution

Physics sir · 3 steps

Same charge, same voltage, so same kinetic energy; the wavelength then goes as one over root mass.

1. Same kinetic energy. Both particles carry charge of magnitude e, so accelerating through 100 V gives each the same kinetic energy K = eV. The momentum is p = √2mK = √2meV.

   λ = h/√2meV ⇒ λₑ/λₚ = √(mₚ/mₑ)

2. Put in masses. mₚ/mₑ = 1.00727/0.00055 ≈ 1831. Its square root is about 42.8.
3. So. λₑ : λₚ ≈ 42.8 : 1, which is nearly 43 : 1, option (d).

The idea to keep. For particles accelerated through the same potential with equal charge, λ ∝ 1/√m. Always check whether it is the speed or the energy that is equal before you compare.

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