Q 12-14-150JEE MainJEE Main 2021 (26 Feb, Shift 1)Easy
LED is constructed from $Ga-As-P$ semiconducting material. The energy gap of this LED is $1.9$ eV. Calculate the wavelength of light emitted and its colour. $h = 6.63\times10^{-34}$ J s and $c = 3\times10^8$ m s$^{-1}$
Answer: (A) $654$ nm and red colour
$$\lambda = \frac{hc}{E_g} = \frac{6.63\times10^{-34}\times3\times10^{8}}{1.9\times1.6\times10^{-19}} \approx 6.54\times10^{-7}\ \text{m} = 654\ \text{nm}$$
This lies in the red part of the visible spectrum.
Solution by Sreeraj P, M.Sc Physics