The de Broglie Wavelength (AQA A Level Physics): Revision Note

A proton and an electron are each accelerated from rest through the same potential difference.

Determine the ratio: $\frac{\mathrm{de}\mathrm{Broglie}\mathrm{wavelength}\mathrm{of}\mathrm{the}\mathrm{proton}}{\mathrm{de}\mathrm{Broglie}\mathrm{wavelength}\mathrm{of}\mathrm{the}\mathrm{electron}}$

• Mass of a proton = 1.67 × 10^–27 kg

• Mass of an electron = 9.11 × 10^–31 kg

[3]

Answer:

Step 1: Determine how the proton and electron can be related via their mass

• The only information given is the mass of the proton and the electron

• When the proton and electron are accelerated through a potential difference, their kinetic energy will increase

• Kinetic energy can therefore be used to relate them via their mass

Step 2: Write out the equation for the de Broglie wavelength in terms of the kinetic energy of the particle

• The de Broglie wavelength

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

• Kinetic energy

$E=\frac{1}{2}m{v}^2=\frac{m{v}^2}{2}=\frac{p^2}{2m}$

• Kinetic energy in terms of momentum

$p=\sqrt{2mE}$

• Substitute the expression for momentum into the de Broglie wavelength equation

$\lambda =\frac{h}{\sqrt{2mE}}$

Step 3: Find the proportional relationship between the de Broglie wavelength and the mass of the particle

$\lambda =\frac{h}{\sqrt{2mE}}=k\frac{1}{\sqrt{m}}$

• Where k is a constant

  • Since h is constant, and E is equal, then:

$\lambda \propto \frac{1}{\sqrt{m}}$ [1 mark]

Step 4: Calculate the ratio

$\frac{{\lambda }_p}{{\lambda }_e}=\frac{1}{\sqrt{m_p}}÷\frac{1}{\sqrt{m_e}}=\frac{1}{\sqrt{m_p}}\times \frac{\sqrt{m_e}}{1}=\frac{\sqrt{m_e}}{\sqrt{m_p}}$ [1 mark]

$\frac{{\lambda }_p}{{\lambda }_e}=\sqrt{\frac{m_e}{m_p}}=\sqrt{\frac{9.11\times {10}^{-31}}{1.67\times {10}^{-27}}}$

$\frac{{\lambda }_p}{{\lambda }_e}=0.0234\left(3\mathrm{s}.\mathrm{f}.\right)$ [1 mark]

• This means that the de Broglie wavelength of the proton is 0.0234 times smaller than that of the electron

• Or that the de Broglie wavelength of the electron is about 40 times larger than that of the proton

Particles with a greater mass, such as a proton, have a greater momentum. The greater the momentum, the smaller the de Broglie wavelength. Always perform a logic check on your answer to check that it makes sense.