Question

What is the de – Broglie wavelength of a proton accelerated through a potential difference of 2 kV?

(a) 0.65 × 10^-13 m

(b) 0.65 × 10^-10 m

(c) 0.65 × 10^-11 m

(d) 0.65 × 10^-20 m

I have been asked this question in class test.

The question is from Wave Nature of Matter in chapter Dual Nature of Radiation and Matter of Physics – Class 12

Answer 1

The correct choice is (b) 0.65 × 10^-10 m

The best explanation: Given: charge of proton = 1.6 × 10^-19; mass of proton = 1.6 × 10^-27; V = 2 kV

λ=\(\frac {h}{p} = \frac {h}{\sqrt {2mE}} = \frac {h}{\sqrt {2mqV}}\)

λ=\(\frac {6.6 \times 10^{-34}}{\sqrt {2\times (1.6 \times 10^{-27})\times (1.6 \times 10^{-19}) \times 2000}}\)

λ=0.65 × 10^-12  m

Therefore, the wavelength is 0.65 × 10^-12 m.