Question

If an LED has to emit $662nm$ wavelength of light, then what should be the band gap energy of its semiconductor? $h = 6.62 \times {10^{ - 34}}Js$

Answer 1

Band energy gap is given by the relation $E = \dfrac{{hc}}{\lambda }$

Complete step by step solution:
Here, the wavelength of the LED $\lambda = 662nm$
$\Rightarrow \lambda = 662 \times {10^{ - 9}}\,m$
Band Energy gap $E = \dfrac{{hc}}{\lambda }$
Where, h is the Planck’s constant whose value is $6.62 \times {10^{ - 34}}Js$ and c is the velocity of light whose value is given by $3 \times {10^8}\,m/s$
Putting the given values in the above equation, we get
$E = \dfrac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{662 \times {{10}^{ - 9}}}} = 3 \times {10^{19}}\,V$

Hence, the band gap energy of its semiconductor is $3 \times {10^{19}}\,V$

Additional Information:
Band gap is a range of energy levels within a given crystal that are impossible for an electron to possess. Generally, a material has several band gaps throughout its band structure. A band gap is the distance between the valence band of electrons and the conduction band. The band gap represents the minimum energy that is required to excite an electron up to a state in the conduction band where it can participate in conduction. If there is a gap between the valence band and the higher energy conduction band, energy must be given for electrons to become free. The size and existence of this band gap gives the difference between conductors, insulators and semiconductors. Semiconductors and insulators may be identified by the size of their band gaps, the former having narrower band gaps and the latter having wider band gaps.

Note: Metals have high electrical conductivity due to their lack of a band gap. There is no band gap separating the valence band in metals from the conduction band. A small fraction of free electrons will always be in the conduction band. This results in a superior electrical conductivity in metals.