In LEED experiments, acceleration voltages of 10 to 200 V are generally used. (a) Calculate the

In LEED experiments, acceleration voltages of 10 to 200 V...

In LEED experiments, acceleration voltages of 10 to $200 \mathrm{V}$ are generally used. $(a)$ Calculate the energies of electrons accelerated by these voltages in $\mathrm{kJ} \mathrm{mol}^{-1}$. ( $b$ ) Calculate the wavelengths of electrons accelerated by these voltages in $\mathrm{nm}$

Key Concepts

De Broglie Wavelength

The de Broglie wavelength is a key concept in quantum mechanics, which describes the wave-like nature of particles. It is given by the relation ? = h/p, where h is Planck's constant and p is the particle's momentum. In the context of accelerated electrons, their kinetic energy (derived from the voltage applied) can be used to compute their momentum, and hence their de Broglie wavelength, which is crucial for interpreting diffraction experiments.

Electron Volt and Unit Conversion

The electron volt (eV) is a standard unit of energy in atomic and molecular physics and is defined as the amount of energy an electron gains when it is accelerated through a potential difference of one volt. Converting this energy into more macroscopic units like kJ/mol involves using known constants such as the charge of the electron and Avogadro's number, thereby linking microscopic measurements to macroscopic energy scales.

Energy Gained by a Charged Particle in an Electric Field

When a charged particle, such as an electron, is accelerated through a potential difference (voltage), it gains kinetic energy equal to the product of its charge and the voltage. This concept, expressed as E = qV, is foundational in understanding how electrons acquire energy in experiments. It underscores that the energy an electron gains depends directly on the voltage applied and its intrinsic charge.

Transcript

00:01 For part a of this problem, we denote the potential energy difference as the greek letter phi.

00:08 And specifically, phi is equivalent to the charge multiplied by the potential.

00:13 So what we can say is that the energy is equivalent to e times phi.

00:21 And so for if the first part of this problem, if we consider phi to be 10 volts, will get that the energy is equivalent to 1 .602 times 10 to the negative 19th kulams multiplied by 10 volts.

00:39 And 10 volts is the same thing as saying joules per kulam.

00:43 So we can get 1 .602 times 10 to the negative 19th kulams multiplied by 10 joules per kulam.

00:52 So our kulams will cancel and we can reduce this to kilojoules.

00:56 And say that our energy is 1 .602 times 10 to the negative 21st kilojoules per electron.

01:06 So now if we consider our voltage to be 200 volts, we can do the exact same thing.

01:13 We can say that our energy is equivalent to 1 .602 times 10 to the negative 19th coulams, multiplied by 200 joules per coulom.

01:24 Our couloms will cancel, and we can simplify this to kill, and say that our energy is 3 .204 times 10 to the negative 20th kilojoules per electron.

01:37 Then we can find the kilojoules per mole by multiplying our energies by avogadro's number.

01:48 So for our first energy, we'll get 1 .62 times 10 to the 21st kilojoules per electron...

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