Messing with an electron

Electricity and Magnetism Level 2

An electron, practically at rest is initially accelerated through a potential difference $100 \text{ V}$ . It then has a de Broglie wavelength of $x_1$ .

It then gets retarded through $19 \text{ V}$ and has a wavelength of $x_2$ .

A final retardation through $32 \text{ V}$ changes the wavelength to $x_3$ .

What is $\displaystyle\frac{x_3-x_2}{x_1}$ as a percentage?

82.3 31.7 74.3 56.6

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Muntasir Sengupta
May 12, 2014

This one depends on the non-relativistic expression for the energy of electrons.

The initial acceleration endows the electron with kinetic energy 100 eV . As a sanity check, classical considerations would predict the speed of such an electron to be $\sqrt{2\times 100\text{ eV}/m_e}\approx 5.93\times10^6$ m/s , well below the relativistic regime.

Therefore, we can say that velocity, and therefore momentum is proportional to $\sqrt{E}$ . The momentum of an electron is given by $\lambda = \frac{h}{p} \sim1/\sqrt{E}$ .

Since we're forming a dimensionless quantity out of all the wavelengths, we can ignore the constants and focus on the $1/\sqrt{E}$ behavior.

When the electron goes through the retarding potentials, it loses kinetic energy.

Therefore, the quantity of interest can be found as

$\displaystyle \frac{x_3-x_2}{x_1} = \frac{1/\sqrt{49 \text{ eV}}-1/\sqrt{81\text{ eV}}}{1/\sqrt{100\text{ eV}}}\approx 31.7$

If I may suggest a small correction to your last line, the ratio itself gives approximately 0.317, since it represents a fractional change. You should include a multiplier of 100% to obtain the final result of 31.7%.

Gregory Ruffa - 7 years ago

thats the same way I thought!

Kartikeya Veeramraju - 7 years, 1 month ago

Messing with an electron

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