Back to Directory
NEET PHYSICSMedium

A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV2.5\text{ eV}. It can detect a signal of wavelength:

A

6000 A˚6000\text{ \AA}

B

4000 nm4000\text{ nm}

C

6000 nm6000\text{ nm}

D

4000 A˚4000\text{ \AA}

Step-by-Step Solution

  1. Condition for Detection: For a photodiode to detect a signal, the energy of the incident photon (EE) must be greater than or equal to the band gap energy (EgE_g) of the semiconductor. That is, E≥EgE \ge E_g.
  2. Calculate Maximum Wavelength: The energy of a photon is related to its wavelength λ\lambda by E=hcλE = \frac{hc}{\lambda}.
  • Substituting hc≈12400 eV⋅A˚hc \approx 12400\text{ eV}\cdot\text{\AA}, we get the maximum detectable wavelength λmax=hcEg=12400 eV⋅A˚2.5 eV=4960 A˚\lambda_{max} = \frac{hc}{E_g} = \frac{12400\text{ eV}\cdot\text{\AA}}{2.5\text{ eV}} = 4960\text{ \AA}.
  1. Evaluate Options: The photodiode can detect any signal with a wavelength less than or equal to 4960 A˚4960\text{ \AA}. 6000 A˚>4960 A˚6000\text{ \AA} > 4960\text{ \AA} (Cannot detect) 4000 nm=40000 A˚>4960 A˚4000\text{ nm} = 40000\text{ \AA} > 4960\text{ \AA} (Cannot detect) 6000 nm=60000 A˚>4960 A˚6000\text{ nm} = 60000\text{ \AA} > 4960\text{ \AA} (Cannot detect) 4000 A˚≤4960 A˚4000\text{ \AA} \le 4960\text{ \AA} (Can detect)
  2. Conclusion: Thus, a signal of wavelength 4000 A˚4000\text{ \AA} can be detected.
Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started