Question

An electron has de broglie wavelength equal to that of a photon. Show that the ratio of the kinetic energy of the electron to the energy of the photon is (m2c4+h2v2)1/2-mc2/hv

Answer 1

Answer on Question#55946 - Physics - Quantum Mechanics

An electron has de Broglie wavelength equal to that of a photon. Show that the ratio of the kinetic energy of the electron to the energy of the photon is $\left(\frac{(m^2c^4 + h^2\nu^2)^{1/2} - mc^2}{h\nu}\right)$ .

Solution:

Since the de Broglie wavelength of the electron is equal to that of a photon, the electron momentum $p$ equals the momentum of photon. Momentum of the photon related to its energy $h\nu$ by the following relation:

$$p c = h \nu$$

Therefore

$$p = \frac {h \nu}{c}$$

The kinetic energy of the electron is given by

$$K = \sqrt {m ^ {2} c ^ {4} + p ^ {2} c ^ {2}} - m c ^ {2}$$

The ratio of the kinetic energy of the electron to the energy of the photon is

$$\frac {K}{h \nu} = \frac {\sqrt {m ^ {2} c ^ {4} + p ^ {2} c ^ {2}} - m c ^ {2}}{h \nu} = \frac {\sqrt {m ^ {2} c ^ {4} + \left(\frac {h \nu}{c}\right) ^ {2} c ^ {2}} - m c ^ {2}}{h \nu} = \frac {\sqrt {m ^ {2} c ^ {4} + h ^ {2} \nu^ {2}} - m c ^ {2}}{h \nu}$$

Answer: $\frac{\sqrt{m^2c^4 + h^2\nu^2} - mc^2}{h\nu}$ .

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