A) 105 V
The De Broglie wavelength of an electron is given by
[tex]\lambda=\frac{h}{p}[/tex]
where
[tex]h=6.63\cdot 10^{-34}Js[/tex] is the Planck constant
p is the electron's momentum
Solving for p and using
[tex]\lambda=0.120 nm=0.12\cdot 10^{-9}m[/tex]
we find
[tex]p=\frac{h}{\lambda}=\frac{6.63\cdot 10^{-34}Js}{0.12\cdot 10^{-9}m}=5.53\cdot 10^{-24} kg m/s[/tex]
The (kinetic) energy of the electron is related to the momentum, p, by the formula
[tex]E=\frac{p^2}{2m}[/tex]
where
m = 9.11×10−31 kg
is the electron's mass
So we calculate the energy of the electron:
[tex]E=\frac{(5.53\cdot 10^{-24} kg m/s)^2}{2(9.11\cdot 10^{-31}kg)}=1.68\cdot 10^{-17} J[/tex]
The energy acquired by an electron moving 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 formula for [tex]\Delta V[/tex] and using the energy of the photon we found previously, we get
[tex]\Delta V=\frac{E}{q}=\frac{1.68\cdot 10^{-17} J}{1.60\cdot 10^{-19}}=105 V[/tex]
B) [tex]1.04\cdot 10^4 V[/tex]
First of all, let's calculate the energy of the x-ray photon, given by:
[tex]E=\frac{hc}{\lambda}[/tex]
where
[tex]h=6.63\cdot 10^{-34}Js[/tex] is the Planck constant
[tex]c=3\cdot 10^8 m/s[/tex] is the speed of light
[tex]\lambda=0.120 nm =0.12\cdot 10^{-9} m[/tex] is the wavelength
Substituting,
[tex]E=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{0.12\cdot 10^{-9} m}=1.66\cdot 10^{-15}J[/tex]
The energy acquired by an electron moving 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 formula for [tex]\Delta V[/tex] and using the energy of the photon we found previously, we get
[tex]\Delta V=\frac{E}{q}=\frac{1.66\cdot 10^{-15} J}{1.60\cdot 10^{-19} C}=1.04\cdot 10^4 V[/tex]
