Question 95: Dual Nature of Matter and Radiation

Question

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

Accepted Answer

To find the value of $$ X $$ in the band gap $$ \left(\frac{\mathrm{X}}{8}ight) \mathrm{eV} $$, 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 $$ 6.6 	imes 10^{-34} $$ JÃÂ·s.
$$ c $$ is the speed of light in a vacuum ($$ \approx 3 	imes 10^{8} $$ 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 $$ e $$ ($$ 1.6 	imes 10^{-19} $$ 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 $$ \lambda_{threshold} $$:
$$ E_{g} = \frac{hc}{\lambda_{threshold}	imes e} $$
$$ E_{g} = \frac{6.6 	imes 10^{-34} 	ext{ JÃÂ·s} \cdot 3 	imes 10^{8} 	ext{ m/s}}{660 	imes 10^{-9} 	ext{ m} 	imes 1.6 	imes 10^{-19} 	ext{ C}} $$
$$ E_{g} = \frac{6.6 	imes 3 	imes 10^{-26}}{660 	imes 1.6} \mathrm{eV} $$
$$ E_{g} = \frac{19.8}{660 	imes 1.6} \mathrm{eV} $$
$$ E_{g} = \frac{19.8}{1056} \mathrm{eV} $$
$$ E_{g} = \frac{18.75}{1000} \mathrm{eV} $$
$$ E_{g} = 1.875 \mathrm{eV} $$
Now, we have to compare this value to $$ \left(\frac{\mathrm{X}}{8}ight) \mathrm{eV} $$ to find the value of $$ X $$:
$$ 1.875 = \frac{X}{8} $$
$$ X = 1.875 	imes 8 $$
$$ X = 15 $$
Therefore, the value of $$ X $$ is 15. The correct answer is Option C.

Answer

11

Answer

13

Answer

15

Answer

21

Dual Nature of Matter and Radiation

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