Electrons moving in a beam have the same de Broglie wavelength as protons in a separate beam moving at a speed of $2.8 \times 10^8 \text{ m s}^{-1}$ - AQA - A-Level Physics - Question 11 - 2017 - Paper 1

To find the speed of the electrons, we can use the de Broglie wavelength formula, which states:

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

where:

• $\lambda$ is the wavelength,
• $h$ is Planck's constant ($6.63 \times 10^{-34} \text{ Js}$),
• $m$ is the mass of the particle (for electrons, approximate mass is $9.11 \times 10^{-31} \text{ kg}$),
• $v$ is the speed of the particle.

Since electrons have the same wavelength as protons, we can say:

$\frac{h}{mv_{electron}} = \frac{h}{mv_{proton}}$

This allows us to set up the equation:

$v_{electron} = v_{proton} \frac{m_{electron}}{m_{proton}}$

Plugging in known values:

• Speed of protons: $v_{proton} = 2.8 \times 10^8 \text{ m s}^{-1}$
• Mass of proton: $m_{proton} \approx 1.67 \times 10^{-27} \text{ kg}$
• Mass of electron: $m_{electron} \approx 9.11 \times 10^{-31} \text{ kg}$

Now we can substitute these values into the equation:

$v_{electron} = 2.8 \times 10^8 \frac{9.11 \times 10^{-31}}{1.67 \times 10^{-27}}$

Calculating this gives:

$v_{electron} \approx 1.5 \times 10^8 \text{ m s}^{-1}$

Thus the speed of the electrons is approximately $1.5 \times 10^8 \text{ m s}^{-1}$, leading us to the correct answer: A.