Question

Q: Calculate the wavelength of light emitted by a semiconductor having band gap energy of 2.8 eV?
A 4430.8 A
8
44.308 A
4.4308 A
O
D 443.08 A​

Answer 1

Given,

Band gap energy=2.8ev

To Find,

Wavelength of light.

Solution,

To find there is a formula,

E=hc/lamda

where,

h=planck's constant

c=speed of light

lambda=wavelength of light emitted

So,

lambda(wavelength)=hc/E

                                 =(6.63×10^-34)×(3×10^8)/(2.8×1.6×10^-19)

                                  =4.4307×10^-7
                                  =4430.7A

Hence, wavelength is 4430.7A