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A p–n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal of wavelength
A
4960 Å
B
6000 Å
C
4000 nm
D
6000 nm
Detailed Solution
A photodiode can detect light only if the photon energy is at least equal to the band gap: $h\nu \geq E_g$, i.e., $\lambda \leq \frac{hc}{E_g}$
Using hc = 12400 eV Å: $\lambda_{max} = \frac{12400}{2.5} = 4960$ Å
So only wavelengths up to 4960 Å (496 nm) can be detected.
6000 Å, 4000 nm (40000 Å) and 6000 nm (60000 Å) are all longer than 4960 Å; their photons have less energy than 2.5 eV and cannot be detected.
Hence it can detect a signal of wavelength 4960 Å.
Note: the source prints this option as '496 Å'; the standard value in the original paper, 4960 Å, is used here.
Using hc = 12400 eV Å: $\lambda_{max} = \frac{12400}{2.5} = 4960$ Å
So only wavelengths up to 4960 Å (496 nm) can be detected.
6000 Å, 4000 nm (40000 Å) and 6000 nm (60000 Å) are all longer than 4960 Å; their photons have less energy than 2.5 eV and cannot be detected.
Hence it can detect a signal of wavelength 4960 Å.
Note: the source prints this option as '496 Å'; the standard value in the original paper, 4960 Å, is used here.