Question

A photon and an electron (mass \(m\)) have the same energy \(E\). The ratio \( \frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} \) of their de Broglie wavelengths is: (\(c\) is the speed of light)

• A. \( c \sqrt{2mE} \)
• B. \( \frac{c \sqrt{2m}}{E} \)
• C. \( c \sqrt{\frac{E}{2m}} \)
• D. \( \sqrt{\frac{E}{2m}} \)

Hint:

Remember the de Broglie wavelength formula \( \lambda = h/p \) and the energy-momentum relations for photons (\(E=pc\)) and non-relativistic particles (\(E=p^2/2m\)).

Answer 1

The Correct Option is C

Step 1: de Broglie wavelength for a particle with momentum
The de Broglie wavelength λ of a particle with momentum p is given by:
λ = h / p, where h is Planck's constant.

Step 2: For a photon
For a photon, energy E = hν = hc / λ_photon, so the momentum of the photon is:
p_photon = E / c.
Therefore, the de Broglie wavelength of the photon is:
λ_photon = h / p_photon = h / (E / c) = hc / E.

Step 3: For an electron
For an electron with mass m and energy E, the kinetic energy is:
E = p_electron² / 2m, so the momentum of the electron is:
p_electron = √(2mE).
Therefore, the de Broglie wavelength of the electron is:
λ_electron = h / p_electron = h / √(2mE).

Step 4: Finding the ratio of the wavelengths
Now, we need to find the ratio λ_photon / λ_electron:
λ_photon / λ_electron = (hc / E) / (h / √(2mE)) = (c √(2mE)) / E = c √(2mE / E²) = c √(2m / E).

Conclusion:
The ratio λ_photon / λ_electron = c √(2m / E).