Question

Calculate the (a) momentum, and (b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.

Answer 1

Answer: (a) Momentum $P =  3.01\times 10^{-26} kg\ m/s$

(b)De Broglie wavelength $\lambda  = 22\times 10^{-9}m$

Explanation:

We know that energy of an atom is

$E = \dfrac{hc}{\lambda }$...............(1)

$E = eV$...............................(2)

Using equation (1) and (2)

$eV = \dfrac{hc}{\lambda }$

$h = 6.626\times 10^{-34} js$,$c=3\times 10^{8} m/s$,$e=1.6\times 10^{-19} c$

So, after putting the values of h,c,e and V,we get

$\lambda  = 22\times 10^{-9}m$

We know the de-brogile relation

$\lambda  = \dfrac{h}{p}$

$P =  3.01\times 10^{-26} kg\ m/s$

(a) Momentum $P =  3.01\times 10^{-26} kg\ m/s$

(b)De Broglie wavelength $\lambda  = 22\times 10^{-9}m$