Page 19 - Photonics Essentials an introduction with experiments

Electrons and Photons

Electrons and Photons 13

Exercise 2.2
Take = 1000 nm = 1 m = 10 –4 cm. For a tungsten light bulb, this is
the wavelength of peak intensity. What is the energy associated with
this wavelength?
Procedure:
f = c, or f = c/

3·10 10 cm/sec
f
10 –4 cm
f = 3·10 /sec . . whew!!!
14
E = 6.6 · 10 –34 × 3·10 –14 = 1.98 · 10 –19 joules

This sounds small, which it is according our everyday scale. Howev-
er, it is very close to the energy that an electron would have if it were
accelerated through a potential of one volt:
1 eV = 1.6 · 10 –19 coul × 1 V = 1.6 · 10 –19 joule
In photonics, the typical energies that you work with involve electrons
in a potential of 1 or 2 V. So we use the energy of an electron acceler-
ated through a potential of 1 V as a handy unit—the electron volt
(eV).
The energy of a photon with a wavelength of 1000 nm (or 1 m) is

1.98 · 10 –19
E = = 1.24 eV (2.8)
1.6·10 –19
It is easy to show that reverse is true. That is, a photon with an en-
ergy of 1 eV has a wavelength of 1.24 m (= 1240 nm). If a photon
with a wavelength of 1 m has an energy of 1.24 eV, what is the ener-
gy of a photon having a wavelength of 0.5 m (= 500 nm)? Answer: E
= 2.48 eV.
What is the energy of red photons ( = 612 nm)? Answer: E = 2.0 eV.

Exercise 2.3
Prove that the energy of any photon is given by
1.24 m
E = eV (2.9)

Prove that the wavelength of any photon is given by

1.24 eV
= m (2.20)
E

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