Answer:
The kinetic energy of an electron in the beam is 5.04 keV.
Explanation:
We need to find the velocity of the electron by using the De Broglie wavelength:
[tex] \lambda = \frac{h}{mv} [/tex]
Where:
λ: is the wavelength = 0.0173 nm
v: is the velocity
m: is the electron's mass = 9.1x10⁻³¹ kg
h: is the Planck constant = 6.62x10⁻³⁴ J.s
[tex] v = \frac{h}{m\lambda} = \frac{6.62 \cdot 10^{-34} J.s}{9.1 \cdot 10^{-31} kg*0.0173 \cdot 10^{-9} m} = 4.21 \cdot 10^{7} m/s [/tex]
Now, we can find the kinetic energy:
[tex] E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}9.1 \cdot 10^{-31} kg*(4.21 \cdot 10^{7} m/s)^{2} = 8.06 \cdot 10^{-16} J*\frac{1 eV}{1.6 \cdot 10^{-19} J} = 5038 eV = 5.04 keV [/tex]
Therefore, the kinetic energy of an electron in the beam is 5.04 keV.
I hope it helps you!
