Light consist of electromagnetic radiations.

Electromagnetic radiation travels in the form of waves. A wave is associated with three characteristic properties, viz, wavelength \((\lambda )\), frequency \((\nu )\) and wavenumber \((\overline { \nu } )\).

**Wavelength\((\lambda ):\)** The distance between two successive crests or troughs is known as wavelength. The units for the measurement of wavelength is nanometre \((nm)\) angstrom (Å), micron (or micro-metre) \((\mu )\) or millimicron \((m\mu )\). These units are inter-related as follows:$$1nm = { 10 }^{ -9 }m = 10 Å\ 1 Å = { 10 }^{ -10 }m = 0.1nm\ 1\mu = { 10 }^{ -6 }m = { 10 }^{ 3 }nm\ $$

**Frequency \((v)\):** The number of wave passing through a given point per second is called the frequency of radiation \((v)\). Frequency has unit of second inverse i.e. \({ s }^{ -1 }\).

Frequency is also expressed in terms of hertz \((Hz)\). **One hertz is equal to one vibration or one oscillation per second.** Frequency and wavelength are related by the equation

\(v = \frac { c }{ \lambda }\)

where \(c\) is the velocity of light \((3 \times { 10 }^{ 8 } m/s)\)

According to Planck’s Quantum theory, the electromagnetic radiation propagates in the space not in continuous manner but in discrete energy packets called quanta. The energy associated with one packet i.e. one quantum is equal to hv, where \(h\) is Planck’s constant \((6.626 \times { 10 }^{ -34 }Js)\) and \(v\) is the frequency or the energy radiation in hertz \(({ s }^{ -1 })\).

Thus, the energy associated with one quantum is Energy in quantum $$E = hv = h\frac { c }{ \lambda } = h\overline { v } c$$ Therefore, the frequency associated with a wave in terms of wave number is given by: $$Wave \quad number \quad \overline { v } = \frac { E }{ hc }$$ The emission or absorption of energy also takes place in discrete instalments of energy i.e. quanta.

**Example: Calculate (i) the frequency in hertz (ii) wavenumber and (iii) wavelength of the radiation in angstrom and nanometre of electromagnetic radiation having energy \(0.1eV\). \((1eV = 1.6 \times { 10 }^{ -19 }J,\) \(h = 6.626 \times { 10 }^{ -34 }Js.\) \(C = 3 \times { 10 }^{ 8 }m{ s }^{ -1 })\)**

**Solution:** Given that,

Energy \(E = 0.1eV\) \(= 0.1 \times \left( 1.6 \times { 10 }^{ -19 }J \right)\)$$ = 1.6 \times { 10 }^{ -20 }J$$from the relation \(\boxed { E = hv = \frac { hc }{ \lambda } = hc\overline { v } }\)

(i) The frequency \(v = \frac { E }{ h }\) \(= \frac { 1.6 \times { 10 }^{ -20 }J }{ 6.626 \times { 10 }^{ -34 }Js }\)$$= 2.415 \times { 10 }^{ 13 }s$$

(ii) Wavenumber \(\overline { v } = \frac { E }{ hc } \) \( = \frac { 1.6 \times { 10 }^{ -20 }J }{ \left( 6.626 \times { 10 }^{ -34 }Js \right) \left( 3 \times { 10 }^{ 8 }m{ s }^{ -1 } \right) }\)\( = 8.049 \times { 10 }^{ 4 }{ m }^{ -1 }\)

(iii) Wavelength \(\lambda = \frac { 1 }{ \overline { v } } \) \( = \frac { 1 }{ 8.049 \times { 10 }^{ 4 }{ m }^{ -1 } }\)

\(= 1.24 \times { 10 }^{ -5 }m\) \(= \frac { 1.24 \times { 10 }^{ -5 }m }{ { 10 }^{ -10 } } Å \) \(= 1.24 \times { 10 }^{ 5 }Å\)

\( = \frac { 1.24 \times { 10 }^{ -5 }m }{ { 10 }^{ -9 } } nm \) \(= 1.24 \times { 10 }^{ 4 }nm\)