BITSAT2017PhysicsDual Nature of MatterActual
The de-Broglie wavelength of a proton (charge = 1 . 6 × 10 - 19 C , m = 1 . 6 × 10 - 27 kg ) accelerated through a potential difference of 1 kV is
Options
- A600   Å
- B0 . 9 × 10 - 12   m
- C7   Å
- D0 . 9   nm
Correct answer
B. 0 . 9 × 10 - 12   m
Step-by-step solution
de-Broglie wavelength, λ = h p = h 2 m E = h 2 m q V ⇒   λ = 6 . 6 × 10 - 34 2 × 1 . 6 × 10 - 27 × 1 . 6 × 10 - 19 × 1000 ⇒   λ = 6 . 6 × 10 - 34 7 . 16 × 10 - 22 = 0 . 9 × 10 - 12   m