To determine the ratio of kinetic energies between an electron and a photon when the electron's de Broglie wavelength is equal to that of a photon, we need to delve into a few concepts in quantum mechanics and energy relations.
Understanding de Broglie Wavelength
The de Broglie wavelength (\(\lambda\)) of a particle is given by the formula:
\(\lambda = \frac{h}{p}\)
where \(h\) is Planck's constant and \(p\) is the momentum of the particle. For an electron with mass \(m\) and velocity \(v\), the momentum can be expressed as:
\(p = mv\)
Therefore, the de Broglie wavelength becomes:
\(\lambda = \frac{h}{mv}\)
Photon Wavelength
For a photon, the wavelength (\(\lambda\)) is related to its energy (\(E\)) by the equation:
\(E = \frac{hc}{\lambda}\)
Here, \(c\) is the speed of light. We can rearrange this to find the wavelength in terms of energy:
\(\lambda = \frac{hc}{E}\)
Setting the Wavelengths Equal
Given that the de Broglie wavelength of the electron equals the wavelength of the photon, we can set the two equations equal:
\(\frac{h}{mv} = \frac{hc}{E}\)
Solving for energy, we can express the energy of the photon:
\(E = \frac{mvc}{h}\)
Kinetic Energy of the Electron
The kinetic energy (\(KE\)) of the electron is given by:
\(KE_{electron} = \frac{1}{2}mv^2\)
Finding the Ratio of Energies
Now, let's find the ratio of the kinetic energies of the electron and the photon. We know:
- The kinetic energy of the photon can be expressed as its energy, \(E_{photon}\).
- From our earlier equations, we can substitute \(E_{photon}\) into the ratio.
Thus, the ratio of the kinetic energy of the electron to the energy of the photon is:
\(\frac{KE_{electron}}{E_{photon}} = \frac{\frac{1}{2}mv^2}{\frac{mvc}{h}} = \frac{h}{2c}\)
Now, the ratio simplifies to:
\(\frac{KE_{electron}}{E_{photon}} = \frac{1}{2}\)
Final Answer
Therefore, the ratio of the kinetic energies of the electron to the photon is:
1 : 2
Thus, the correct answer is option B: 1 : 2.
