Question

You use an electron microscope in which the matter wave associated with the electron beam has a wavelength of 0.0173 nm. What is the kinetic energy of an electron in the beam, expressed in electron volts?

Answer 1

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:

$\lambda = \frac{h}{mv}$

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

$v = \frac{h}{m\lambda} = \frac{6.62 \cdot 10^{-34} J.s}{9.1 \cdot 10^{-31} kg*0.0173 \cdot 10^{-9} m} = 4.21 \cdot 10^{7} m/s$

Now, we can find the kinetic energy:

$E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}9.1 \cdot 10^{-31} kg*(4.21 \cdot 10^{7} m/s)^{2} = 8.06 \cdot 10^{-16} J*\frac{1 eV}{1.6 \cdot 10^{-19} J} = 5038 eV = 5.04 keV$

Therefore, the kinetic energy of an electron in the beam is 5.04 keV.

I hope it helps you!