Where the simple formula stops working
De Broglie's relation \(\lambda = h/p\) holds at every speed. What changes at high speed is the link between the momentum and the kinetic energy. Lesson 1 used \(p = \sqrt{2mK}\), which comes from Newton's \(K = p^2/2m\). Lesson 5 replaced that with the energy-momentum relation. For an electron in an electron microscope, the difference matters.
\[ E = K + m_0c^2 \]
Why. the total energy is the kinetic energy plus the rest energy.
\[ (pc)^2 = E^2 - (m_0c^2)^2 \]
Why. the energy-momentum relation of lesson 5.
\[ (pc)^2 = K^2 + 2Km_0c^2 \]
Why. put in \(E = K + m_0c^2\) and expand the square; the \((m_0c^2)^2\) terms cancel.
\[ pc = \sqrt{K^2 + 2Km_0c^2} \]
Read it. take the square root. When \(K \ll m_0c^2\), the \(K^2\) term is negligible and \(p = \sqrt{2m_0K}\), as before.
An electron microscope accelerates electrons from rest through \(100\,\text{kV}\). Find their de Broglie wavelength with the exact formula, and compare it with \(1.227/\sqrt{V}\,\text{nm}\).
The kinetic energy is \(K = 0.100\,\text{MeV}\), and the rest energy is \(0.51\,\text{MeV}\).
\(pc = \sqrt{0.100^2 + 2 \times 0.100 \times 0.51} = \sqrt{0.112} = 0.335\,\text{MeV}\).
\(\lambda = hc/pc = 1240\,\text{eV nm} / 335\,000\,\text{eV} = 0.0037\,\text{nm}\).
The simple formula gives \(1.227/\sqrt{100\,000} = 0.0039\,\text{nm}\), about \(5\%\) too long.
Values used: \(hc = 1240\,\text{eV nm}\), the electron's rest energy \(m_0c^2 = 0.51\,\text{MeV}\), \(h = 6.63 \times 10^{-34}\,\text{J s}\).
A one-line test
Divide the simple wavelength by the exact one. The ratio is \(\sqrt{1 + K/2m_0c^2}\), so the error depends only on \(K\) compared with the rest energy. For electrons, with \(2m_0c^2 = 1.02\,\text{MeV}\):
- at \(10\,\text{kV}\) the simple formula is \(0.5\%\) too long;
- at \(50\,\text{kV}\), \(2.4\%\) too long;
- at \(100\,\text{kV}\), \(4.8\%\) too long;
- at \(1\,\text{MV}\), \(41\%\) too long.
So for electrons, \(h/\sqrt{2mK}\) is good to a few per cent below about \(50\,\text{kV}\), and the exact formula is needed above it. A proton has a rest energy nearly two thousand times larger. The simple formula serves it up to tens of MeV.
An electron is accelerated from rest through \(1\,\text{MV}\), so \(K = 1\,\text{MeV}\). Take \(m_0c^2 = 0.51\,\text{MeV}\) and \(hc = 1240\,\text{eV nm}\). Find its de Broglie wavelength, in units of \(10^{-3}\,\text{nm}\).
For which of these is \(\lambda = h/\sqrt{2mK}\) accurate to within about \(3\%\)?
Same wavelength, same momentum, different energy
Give a photon and an electron the same wavelength \(\lambda\). Both then have the same momentum \(p = h/\lambda\). Their energies are not the same. The photon has no rest mass, so its energy is \(E = pc = hc/\lambda\). A slow electron has kinetic energy \(K = p^2/2m_0 = (pc)^2/2m_0c^2\). The ratio of the two is \(E/K = 2m_0c^2/pc\).
A photon and an electron each have a wavelength of \(1\,\text{nm}\). Find the momentum of each, the energy of the photon and the kinetic energy of the electron.
Both have \(p = h/\lambda = 6.63 \times 10^{-34} / 10^{-9} = 6.63 \times 10^{-25}\,\text{kg m/s}\).
The photon: \(E = hc/\lambda = 1240\,\text{eV nm} / 1\,\text{nm} = 1240\,\text{eV}\).
The electron: \(K = (pc)^2/2m_0c^2 = 1240^2 / (1.02 \times 10^6) = 1.5\,\text{eV}\).
The photon carries about \(820\) times as much energy as the electron with the same wavelength.
Fig 6.1 joins the two halves of the chapter. At \(0.1\,\text{nm}\) the photon, an X-ray, has \(12.4\,\text{keV}\), and the electron has \(151\,\text{eV}\), about \(80\) times less. As the wavelength shrinks, \(pc\) grows towards the electron's rest energy, and the electron becomes fast. At \(0.0037\,\text{nm}\), the wavelength of the microscope's electrons, the photon has \(335\,\text{keV}\) and the electron \(100\,\text{keV}\). At still shorter wavelengths \(pc\) dwarfs the rest energy, and both energies approach \(pc\).
An electron has a de Broglie wavelength of \(1\,\text{nm}\). Find its kinetic energy in eV. Take \(hc = 1240\,\text{eV nm}\) and \(m_0c^2 = 0.51\,\text{MeV}\).
A photon has a wavelength of \(0.1\,\text{nm}\). Find its energy in keV. Take \(hc = 1240\,\text{eV nm}\).
A photon and a slow electron have the same wavelength. Which statement is true?
Estimate the de Broglie wavelength of a student of mass \(60\,\text{kg}\) walking at \(1.5\,\text{m/s}\). Give the answer in units of \(10^{-36}\,\text{m}\).
- Using \(K = p^2/2m\) for an electron of hundreds of keV.Compare \(K\) with the rest energy \(0.51\,\text{MeV}\). Near or above it, use \((pc)^2 = K^2 + 2Km_0c^2\).
- Using \(E = hc/\lambda\) for an electron.\(hc/\lambda = pc\) is the energy only of a particle with no rest mass. An electron's kinetic energy is smaller.
- Thinking equal wavelengths mean equal speeds.Equal wavelengths mean equal momenta. A photon moves at \(c\); an electron of the same momentum moves far slower.