Question

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

Answer 1

3.064x10^-6 V

Step-by-step explanation:

Detailed explanation and calculation is shown in the image below.

Answer 2

3.071*10^{-6}V

Step-by-step explanation:

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

`E_k=(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=(h)/(\lambda)\\\\v=(h)/(m\lambda)`

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

wavelength = 700*10^{-9}m

`v=(6.62*10^(-34)Js)/((9.1*10^(-31)kg)(700*10^(-9)m))=1039.24(m)/(s)`

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

`V=(m_ev^2)/(2e)=((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