Question

a beam of electrons (m = 9.11 x 10-31 kg/electron) has an average speed of 1.5 x 108 m/s. what is the wavelength of electrons having this average speed, in picometers?

Explanation:

Step1: Calculate the momentum of the electron

The momentum \(p = mv\), where \(m = 9.11\times10^{-31}\text{ kg}\) (mass of electron) and \(v = 1.5\times 10^{8}\text{ m/s}\) (speed of electron).

$$p=(9.11\times 10^{-31}\text{ kg})\times(1.5\times 10^{8}\text{ m/s}) = 1.3665\times 10^{-22}\text{ kg}\cdot\text{m/s}$$

Step2: Use de - Broglie wavelength formula

The de - Broglie wavelength formula is \(\lambda=\frac{h}{p}\), where \(h = 6.626\times 10^{-34}\text{ J}\cdot\text{s}\) (Planck's constant).

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Step3: Convert meters to picometers

Since \(1\text{ m}=10^{12}\text{ pm}\), then \(\lambda=(4.85\times 10^{-12}\text{ m})\times(10^{12}\text{ pm/m})\)

Answer:

\(4.85\)