Question

Through what potential difference must an electron be accelerated from rest to have a de broglie wavelength of 700 nm ?

Answer 1

Answer:

3.071*10^{-6}V

Explanation:

To find the potential difference you take into account the kinetic energy of the electron generated by the potential:

$E_k=\frac{1}{2}m_ev^2=eV$  (1)

m: mass of the electron = 9.1*10^{-31}kg

v: velocity of electron

V: potential difference

e: charge of electron = 1.6*10^{-19}C

Thus, is necessary to find the velocity. By using the Broglie's relation you obtain:

$p=m_ev=\frac{h}{\lambda}\\\\v=\frac{h}{m\lambda}$

h: Planck's constant = 6.62*10^{-34}Js

wavelength = 700*10^{-9}m

$v=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(700*10^{-9}m)}=1039.24\frac{m}{s}$

By doing V the subject of the formula (1) you obtain:

$V=\frac{m_ev^2}{2e}=\frac{(9.1*10^{-31}kg)(1039.24m/s)^2}{2(1.6*10^{-19}C)}=3.071*10^{-6}V=3.071\mu V$

the potential difference required is 3.071*10^{-6}V

Answer 2

Answer:

3.064x10^-6 V

Explanation:

Detailed explanation and calculation is shown in the image below.