Question

A proton is accelerated through 225 V. Its de-broglie wavelength is

1- 0.1nm

2-0.2nm

3-0.3 nm

4-0.4nm

Answer 1

Debroglie wavelength is given by:λ = h2mqV√m is the mas of the proton = 1.6×10−27 kgq is the charge on proton = 1.6×10−19 CV = 225 voltλ = 6.6×10−342×1.6×10−27×1.6×10−19×225√ λ = 0.194 ×10−11 mλ = 0.001 nm