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Question

(D) The energy of a photon required to excite an electron across the band gap is given by $E = h
u$,where $E$ is the band gap energy,$h$ is Planck's

Vedclass · Accepted Answer

$5	imes10^{14}\, Hz$

Vedclass · Answer

(D) The energy of a photon required to excite an electron across the band gap is given by $E = h
u$,where $E$ is the band gap energy,$h$ is Planck's constant,and $
u$ is the frequency.Given,$E = 2.0\, eV = 2.0 	imes 1.6 	imes 10^{-19}\, J = 3.2 	imes 10^{-19}\, J$.Planck's constant $h \approx 6.63 	imes 10^{-34}\, J\cdot s$.The minimum frequency $
u$ is given by $
u = E / h$.$
u = (3.2 	imes 10^{-19}) / (6.63 	imes 10^{-34}) \approx 0.4826 	imes 10^{15}\, Hz$.$
u \approx 4.83 	imes 10^{14}\, Hz$.Rounding to the nearest option,we get $
u \approx 5 	imes 10^{14}\, Hz$.

The photodiode is made of a material with a band gap of $2.0\, eV$. The minimum frequency of the radiation that can be absorbed by the material is nearly

Similar Questions

Assertion: $A$ $P-N$ junction diode is used in reverse bias to detect the intensity of light.
Reason: In reverse bias condition,current is small but it is more sensitive to changes in intensity of incident light.

Explain the construction and working of a photodiode.

To measure light intensity,we use:

If a semiconductor photodiode can detect a photon with a maximum wavelength of $400\, nm$, then its band gap energy is (Given: Planck's constant $h = 6.63 \times 10^{-34}\, J \cdot s$, Speed of light $c = 3 \times 10^{8}\, m/s$) (in $ eV$)

$A$ $p-n$ junction is fabricated from a semiconductor with a band gap of $2.8 \ eV$. What approximate wavelength can it $NOT$ detect (in $nm$)?
[Use $h = 6 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$]

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