Question 11.13: What is the
(b) speed, and
(c) de Broglie wavelength of an electron with the kinetic energy of 120 eV.
For the electron, we can write the relation for kinetic energy as,
Where, v = Speed of the electron
Step 2: Find the momentum of the electron.
Momentum of the electron, p = mv = 9.1 x x 6.496 x
= 5.91 x kg ms
Therefore, the momentum of the electron is 5.91x kg m
Speed of the electron, v = 6.496 x m/s [from part (a)]
Step: Find de Broglie wavelength of an electron.
De Broglie wavelength of an electron having a momentum P is given as:
Therefore, the de Broglie wavelength of the electron is 0.112 nm.