Question

the KE of an electron is 6.625 * 10 ^-28 J. what is the frequency of the de broglie wave associated with it

Answer 1

Answer:

The wavelength of a 2 eV photon is given by:

l = h c / Eph = 6.625 x 10-34 x 3 x 108/(1.6 x 10-19 x 2) = 621 nm.

where the photon energy was multiplied with the electronic charge to convert the energy in Joule rather than electron Volt.

The kinetic energy of an electron is related to its momentum by:

T = p2/2m

from which we find the momentum, p:

p = (2mT)1/2 = (2 x 9.1 x 10-31 x 1.6 x 10-19 x 2)1/2 = 7.63 x 10-25 kg m/s.

The de Broglie wavelength of the electron is then obtained from:

l = h/p = 6.625 x 10-34 / 7.63 x 10-25 = 0.87 nm

Explanation:

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