The de-Broglie wavelength is related to the momentum of a particle by the equation:
\[
\lambda = \frac{h}{p}
\]
where \( \lambda \) is the de-Broglie wavelength, \( h \) is Planck's constant, and \( p \) is the momentum of the particle.
For a photon, the momentum \( p \) is related to its energy \( E \) by the equation:
\[
p = \frac{E}{c}
\]
where \( E \) is the energy of the photon and \( c \) is the speed of light.
For an electron, the momentum \( p \) is given by:
\[
p = \frac{h}{\lambda}
\]
where \( \lambda \) is the de-Broglie wavelength of the electron.
Since the electron and photon are given to have the same de-Broglie wavelength, their momentum must also be the same.
- Velocity: The velocities of the electron and photon are not necessarily the same, since their masses are different (photon has no rest mass while the electron does).
- Momentum: The electron and photon have the same de-Broglie wavelength, hence the same momentum.
- Angular momentum: Angular momentum depends on the specific motion of the particle and is not directly related to the de-Broglie wavelength.
- Energy: The energy of the photon and electron are not necessarily the same, as their energies depend on different factors (photon's energy depends on frequency, electron's energy depends on its kinetic energy).
Thus, the correct answer is momentum.