Calculate the
(a) momentum, and
(b) de-Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Given,
Wavelength of light, λ = 488 nm = 488 x 10–9 m
Stopping potential, V0 = 0.38V
As,
Given,
Frequency of light, = 7.21 x 1014 Hz
Maximum speed of electrons, vmax = 6.0 x 105 ms–1
Mass of the electron, m = 9 x 10–31 kg
Applying Einstein's photoelectric equation,
What is the
(a) momentum,
(b) speed, and
(c) de-Broglie wavelength of an electron with kinetic energy of 120 eV.
given,
Kinetic energy of the electron, K.E = 120 eV
(a) Momentum of the electron is given as,
(b) Now, since
we have,
(c) De- broglie wavelength of electron is given by,