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Question: An LED is constructed from a pn junction based on a certain semi- conducting material whose energy g...

An LED is constructed from a pn junction based on a certain semi- conducting material whose energy gap is 1.9eV1.9eV. Then the wavelength of the emitted light is:
A. 2.9×10−9m2.9\times {{10}^{-9}}m
B. 1.6×10−8m1.6\times {{10}^{-8}}m
C. 6.5×10−7m6.5\times {{10}^{-7}}m
D. 9.1×10−5m9.1\times {{10}^{-5}}m

Explanation

Solution

Hint: Light is produced in an LED when an electron falls back to its base state (valence band) after getting excited by the current flowing through it, to the conduction band. The energy gap between the conduction band and the valence band is released by the electron in the form of the light that we see coming from the LED.

Formula used:
Energy EE of a radiation of wavelengthλ\lambda is given by
E=hcλE=\dfrac{hc}{\lambda }
where h=6.626×10−34J.sh=6.626\times {{10}^{-34}}J.s is the Planck’s constant and c=3×108m/sc=3\times {{10}^{8}}m/s is the speed of light.
1eV=1.6×10−19C×1V=1.6×10−19J1eV=1.6\times {{10}^{-19}}C\times 1V=1.6\times {{10}^{-19}}J

Complete step by step answer:
An LED is a pn junction diode semiconductor device that is used to produce light. When electric current flows in the LED, the electrons in the conduction band (base state) of the diode get excited and jump up to the valence band (excited state) by crossing the band energy gap between the two bands by virtue of the energy provided by the current.
When the electron falls back to its base state, this same amount of energy (that is the energy of the band gap) is released by the electron in the form of radiation. LEDs are manufactured in such a way that the band gap corresponds to energy carried by radiation in the visible light range. By using different materials for the semiconductor, different band gaps and hence, different colours of light can be achieved.
Now, let us analyze the question.
The given band gap energy is E=1.9eVE=1.9eV.
Now, since,1eV=1.6×10−19C×1V=1.6×10−19J1eV=1.6\times {{10}^{-19}}C\times 1V=1.6\times {{10}^{-19}}J,
E=1.9×1.6×10−19J=3.04×10−19JE=1.9\times 1.6\times {{10}^{-19}}J=3.04\times {{10}^{-19}}J --(1)
This amount of energy is carried by the radiation released by the electron.
Now, Energy EE of a radiation of wavelengthλ\lambda is given by
E=hcλE=\dfrac{hc}{\lambda } --(2)
where h=6.626×10−34J.sh=6.626\times {{10}^{-34}}J.s is the Planck’s constant and c=3×108m/sc=3\times {{10}^{8}}m/s is the speed of light.
Therefore, putting (1) in (2), we get,
3.04×10−19=6.626×10−34×3×108λ3.04\times {{10}^{-19}}=\dfrac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{\lambda }
∴λ=6.626×10−34×3×1083.04×10−19=6.54×10−7≈6.5×10−7m\therefore \lambda =\dfrac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{3.04\times {{10}^{-19}}}=6.54\times {{10}^{-7}}\approx 6.5\times {{10}^{-7}}m
Hence, the wavelength of the emitted light is 6.5×10−7m6.5\times {{10}^{-7}}m.
Therefore, the correct option is C)6.5×10−7m6.5\times {{10}^{-7}}m.

Note: One common mistake in such types of questions is that students forget to convert the energy in eVeV toJJ. This causes a big error in the answer. Students must be careful of the units that the energy in the question has been given in. Sometimes questions are purposefully set having these energy units just to test the students’ awareness and knowledge of converting the units.
Students can also verify whether their answer seems correct by applying logic and common sense. Since, LEDs are supposed to emit radiation in the visible range spectrum, therefore automatically, even before solving the question it is quite evident that option C) is the correct answer since the other options don’t even match the wavelength for the visible spectrum range.