Conductivity of a photodiode starts changing only if the wavelength of incident light is less than 660 nm. The band gap of photodiode is found to be $$\frac{X}{8}$$ eV. The value of X is: (Given $$h = 6.6 \times 10^{-34}$$ J s, $$e = 1.6 \times 10^{-19}$$ C)

The conductivity of the photodiode changes when the wavelength of incident light is less than or equal to 660 nm. This wavelength corresponds to the minimum photon energy required to excite electrons across the band gap, which equals the band gap energy $$E_g$$.

The energy of a photon is given by:

$$E = \frac{hc}{\lambda}$$

where:

• $$h = 6.6 \times 10^{-34} \, \text{J s}$$ (Planck's constant),
• $$c = 3 \times 10^8 \, \text{m/s}$$ (speed of light),
• $$\lambda = 660 \, \text{nm} = 660 \times 10^{-9} \, \text{m} = 6.6 \times 10^{-7} \, \text{m}$$ (wavelength).

Substitute the values:

$$E = \frac{(6.6 \times 10^{-34}) \times (3 \times 10^8)}{6.6 \times 10^{-7}}$$

Simplify the expression:

Numerator: $$6.6 \times 10^{-34} \times 3 \times 10^8 = 19.8 \times 10^{-26} = 1.98 \times 10^{-25}$$

Denominator: $$6.6 \times 10^{-7}$$

So,

$$E = \frac{1.98 \times 10^{-25}}{6.6 \times 10^{-7}} = \frac{1.98}{6.6} \times 10^{-25 + 7} = 0.3 \times 10^{-18} = 3 \times 10^{-19} \, \text{J}$$

This energy is in joules, but the band gap is given in electron volts (eV). The conversion factor is $$1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$$. Therefore, convert energy to eV:

$$E_g = \frac{3 \times 10^{-19}}{1.6 \times 10^{-19}} = \frac{3}{1.6} = 1.875 \, \text{eV}$$

The problem states that the band gap is $$\frac{X}{8} \, \text{eV}$$. So,

$$\frac{X}{8} = 1.875$$

Solving for $$X$$:

$$X = 1.875 \times 8 = 15$$

Thus, the value of $$X$$ is 15.