QJEEM24Phd274: JEE Main 2024 Physics - Class 12 JEEM24Phd274 Solution: Conductivity of a photodiode starts changing only if the…

To find the value of $X$ in the band gap $\left(\frac{\mathrm{X}}{8}\right) \mathrm{eV}$ , we need to understand the relationship between the wavelength of light that can result in changes in the conductivity of a photodiode and the photodiode's band gap energy.

The band gap energy ( $E_{g}$ ) of a material is the minimum energy required for an electron to transition from the valence band to the conduction band, thus creating a hole-electron pair and allowing conductivity to occur. When light with a certain wavelength ( $\lambda$ ) is incident upon the photodiode, if the energy of the photons is greater than or equal to the band gap energy of the photodiode, then the photons can excite electrons and change the conductivity of the photodiode.

The energy of a photon ( $E_{photon}$ ) is given by the equation:

$E_{photon} = \frac{hc}{\lambda}$

where:

• $h$ is Planck's constant, which is given as $6.6 \times 10^{-34}$ J·s.
• $c$ is the speed of light in a vacuum ( $\approx 3 \times 10^{8}$ m/s).
• $\lambda$ is the wavelength of the incident light.

Then the band gap energy in terms of electron volts is found by converting the energy from joules to electron volts (eV) using the charge of an electron $e$ ( $1.6 \times 10^{-19}$ C):

$E_{g} = \frac{hc}{\lambda e}$

Given that the photodiode starts conducting when the wavelength of light is less than $660 \mathrm{~nm}$ , we should use this wavelength as the threshold wavelength $\lambda_{threshold}$ :

$E_{g} = \frac{hc}{\lambda_{threshold}\times e}$

$$E_g=\frac{6.6\times {10}^{-34}\text{ J·s}\cdot 3\times {10}^8\text{ m/s}}{660\times {10}^{-9}\text{ m}\times 1.6\times {10}^{-19}\text{ C}}$$

$E_{g} = \frac{6.6 \times 3 \times 10^{-26}}{660 \times 1.6} \mathrm{eV}$

$E_{g} = \frac{19.8}{660 \times 1.6} \mathrm{eV}$

$E_{g} = \frac{19.8}{1056} \mathrm{eV}$

$E_{g} = \frac{18.75}{1000} \mathrm{eV}$

$E_{g} = 1.875 \mathrm{eV}$

Now, we have to compare this value to $\left(\frac{\mathrm{X}}{8}\right) \mathrm{eV}$ to find the value of $X$ :

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

$X = 1.875 \times 8$

$X = 15$

Therefore, the value of $X$ is 15. The correct answer is Option C.