Question

An electron has a deBroglie 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
hv
m c h v mc
( 2 4 + 2 2)1/ 2 − 2

Answer 1

Answer on Question #51442, Physics, Solid State Physics

An electron has a deBroglie 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

$$\frac {\left[ \left(m ^ {2} c ^ {4} + h ^ {2} v ^ {2}\right) ^ {1 / 2} - m c ^ {2} \right]}{h v}$$

Solution:

The de Broglie wavelength for electron is given by Eq.(1)

$$\lambda = \frac {h}{m u}$$

where $h$ is the Planck constant; $m$ is the mass of electron; $u$ is the velocity of electron.

The wavelength for photon

$$\lambda = c / \nu$$

where $c$ is the velocity of light.

Thus,

$$v = \frac {h \nu}{m c}$$

The kinetic energy

$$E _ {K} = \frac {1}{2} m u ^ {2} = \frac {m}{2} \left[ \frac {h \nu}{m c} \right] ^ {2}$$

The kinetic energy for electron

$$\begin{array}{l} E _ {e} = \frac {h ^ {2} v ^ {2}}{2 m c ^ {2}} = \frac {h ^ {2} v ^ {2}}{2 m c ^ {2}} + m c ^ {2} - m c ^ {2} = m c ^ {2} \left[ 1 + \frac {h ^ {2} v ^ {2}}{2 m ^ {2} c ^ {4}} \right] - m c ^ {2} \approx \\ \approx m c ^ {2} \left[ 1 + \frac {h ^ {2} v ^ {2}}{m ^ {2} c ^ {4}} \right] ^ {1 / 2} - m c ^ {2} = \left(m ^ {2} c ^ {4} + h ^ {2} v ^ {2}\right) ^ {1 / 2} - m c ^ {2} \\ \end{array}$$

For photon

$$E _ {p} = h v$$

So,

$$\frac {E _ {p}}{E _ {p}} = \frac {\left(m ^ {2} c ^ {4} + h ^ {2} v ^ {2}\right) ^ {1 / 2} - m c ^ {2}}{h v}$$

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