Question

An electron is accelerated through a potential difference of 10,000 V. its de broglie wavelength is nearly: (me=9*10^-31kg)

Answer 1

potential difference , V = 10000 volts

charge on electron , q = 1.6 × 10^-19 C

mass of electron, m = 9.1 × 10^-31 Kg

a/c to De-broglie's wavelength,

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

where h is plank's constant. i.e., h = 6.64 × 10^-34 Js

so, wavelength = 6.64 × 10^-34/√(2 × 1.6 × 10^-19 × 10000 × 9.1 × 10^-31)

= 6.64 × 10^-34/(5.4 × 10^-23) m

= (6.64/5.4) × 10^(-34 + 23) m

= 1.2296 × 10^-11 m

= 12.296 × 10^-12 m or 12.296 pm

hence, De-broglie's wavelength is 12.296 pm

Answer 2

Answer:

12.2*10^-12

Explanation

K = eV

K = 1/2 mv^2 == p^2/2m

P = [2mKE]^1/2 == [2meV]^1/2

the de broglie wavelength = h/p

wavelength = h/[2meV]^1/2

substituting the numericals for h,m,e...we get,

wavelength = 1.227/[v]^1/2 nm

therefore wavelength = 1.227/[10000]^1/2 nm

                               = 1.227/100 nm

                                = 1.227 * 10^-2 *10^-9 m

                                =1.227 * 10^-11 m

                                 =12.27 *10^-12m

hope u guys understood...