Question

The de-broglie wavelength of a proton (mass=1.61027kg) accelerated through a potential difference of 1 kv is

Answer 1

according to De- Broglie's wavelength equation,

$\boxed{\bf{\lambda=\frac{h}{\sqrt{2qVm}}}}$

where , $\lambda$ denotes wavelength, q denotes charge , m denotes mass and V denotes voltage.

here, h = 6.6 × 10^-34 J.s , q = 1.6 × 10^-19 C , m = 1.6 × 10^-27 kg and V = 1kv = 1000volts

so, $\lambda$ = 6.6 × 10^-34/√{2 × 1.6 × 10^-19 × 1.6 × 10^-27 × 1000}

$\lambda$ = 6.6 × 10^-34/3.2 × 10^-22 × √5

= 3.3 × 10^-12/1.6 × √5 m

= 0.922 × 10^-12 m