P-N Junction Diode and its applications: Practice Problem & Solution

Assertion (A): The semiconductor used for fabrication of visible LED must at least have a band gap of 1.8 eV. Reason (R): The spectral range of visible light is from 0.4 eV to 1.8 eV.

Both (A) & (R) are true and the (R) is the correct explanation of the (A)

Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

(A) is true but (R) is false

Both (A) and (R) are false

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of P-N Junction Diode and its applications. Understanding the underlying formula is key to arriving at the correct answer below:

For visible light emission, the photon energy must be within the visible spectrum, which corresponds to energies from approximately \(1.8\) to \(3.1 \text{ eV}\). Thus, Assertion (A) is true. Reason (R) states the range from \(0.4\) to \(1.8 \text{ eV}\), which is incorrect for visible light.

P-N Junction Diode and its applications: Practice Problem & Solution

Short Solution (Best for NEET):

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