Answer:
66.3 V
Explanation:
The wavelength of the electron must be equal to that of the x-ray photon:
[tex]\lambda=0.150 nm=0.15\cdot 10^{-9}m[/tex]
the De Broglie wavelength of the electron is related to its momentum, p, by the formula
[tex]p=\frac{h}{\lambda}[/tex]
where h is the Planck constant. Solving the formula, we find
[tex]p=\frac{6.63\cdot 10^{-34} Js}{0.15\cdot 10^{-9}m}=4.4\cdot 10^{-24} kg m/s[/tex]
Now we can find the electron's energy using the formula
[tex]E=\frac{p^2}{2m}=\frac{(4.4\cdot 10^{-24} kg m/s)^2)}{2(9.11\cdot 10^{-31} kg)}=1.06\cdot 10^{-17} J[/tex]
Then, we know that the energy of an electron accelerated through a potential difference of [tex]\Delta V[/tex] is
[tex]E=q\Delta V[/tex]
where
[tex]q=1.60\cdot 10^{-19} C[/tex] is the electron charge
Solving the equation for the potential difference, we find
[tex]\Delta V=\frac{E}{q}=\frac{1.06\cdot 10^{-17} J}{1.60\cdot 10^{-19} C}=66.3 V[/tex]
