Question

Through what potential difference ΔV must electrons be accelerated (from rest) so that they will have the same wavelength as an x-ray of wavelength 0.150 nm ? Use 6.63×10−34 J⋅s for Planck's constant, 9.11×10−31 kg for the mass of an electron, and 1.60×10−19 C for the charge on an electron. Express your answer using three significant figures.

Answer 1

66.3 V

Step-by-step explanation:

The wavelength of the electron must be equal to that of the x-ray photon:

`\lambda=0.150 nm=0.15\cdot 10^(-9)m`

the De Broglie wavelength of the electron is related to its momentum, p, by the formula

`p=(h)/(\lambda)`

where h is the Planck constant. Solving the formula, we find

`p=(6.63\cdot 10^(-34) Js)/(0.15\cdot 10^(-9)m)=4.4\cdot 10^(-24) kg m/s`

Now we can find the electron's energy using the formula

`E=(p^2)/(2m)=((4.4\cdot 10^(-24) kg m/s)^2))/(2(9.11\cdot 10^(-31) kg))=1.06\cdot 10^(-17) J`

Then, we know that the energy of an electron accelerated through a potential difference of
`\Delta V` is

`E=q\Delta V`

where

`q=1.60\cdot 10^(-19) C` is the electron charge

Solving the equation for the potential difference, we find

`\Delta V=(E)/(q)=(1.06\cdot 10^(-17) J)/(1.60\cdot 10^(-19) C)=66.3 V`