The solar spectrum has a spectral mismatchwith the band gap of the absorber material ofa single junction solar cell. Area's I, II, III andIV in the figure above correspond to differentmechanisms in which the energy of the lightis utilized. Which of the following statementsis FALSE?0.90.80.70.6IIIV0.50.40.30.2II0.10.511.522.5Bandgap energy (eV)The solar spectrum has a spectral mismatch with the band gap of the absorber material of a single junctionsolar cell. Area's I, II, Il and IV in the figure above correspond to different mechanisms in which the energy ofthe light is utilized. Which of the following statements is FALSE?Area IV corresponds to losses of photons with energies smaller than the band gapArea Il corresponds to heat losses related to the excess energy of the absorbed photons.For multi-junction solar cells areas Il and IV are reduced in reference to Figure 1.Area I corresponds to the maximum energy of the light which can be converted in to electricityincluding the losses of charge carriers recombination.Area III corresponds to the energy used to excite an electron from the valence to conduction bandFraction of incidentphoton energy3.

Given that,The solar spectrum has a spectral mismatch with the band gap of the absorber material of…

Answered: The solar spectrum has a spectral…

Similar questions

An empirical interatomic pair potential for xenon atoms, in units of eV, and nm, and the lattice parameter of Xenon is equal to 0.630 nm. V(r) = 12.6 X 10-7 31.8 X 10-4 712 The calculated interatomic separation (nearest-neighbour) distance is: A. 0.2227 nm. B. 0.3536 nm C. 0.4455 nm D. 1.414 nm
Which of the following statements is NOT true about Fermi Resonance? Shifting of the energies and intensities of absorption bands in an infrared causes a spilt and shift in intensity of peaks with similar energies It results from coupling of a fundamental vibration with an overtones used to help explain and assign peaks in vibrational spectra that which correspond to specific fundamental vibrations
The minority carriers in N-type Si, having a conductivity of (0.01(2.m)-1}, and mobility of electron=D0.36 (m2/(v.sec)).(ni=1.45*10^10 /m3) is O 1.74 x 1017/m3 O 17.36*10^16 /m3 O 1211.117512 /m3 O 18.45*10^4 /m3 O Other:
Graph below shows the electron occupancy probability P(E) as a function of energy for Bismuth (mBi = 3.47 × 10-25 kg) at the temperature T = 0 K. What is the number of conduction electrons per unit volume for Bismuth? 1 1 2 3 4 5 6 7 8 E (ev) P(E)
Figure 1 shows the absorption coefficient for several direct and indirect bandgap semiconductors. Analyses these figure in term of indirect bandgap materials. ABSORPTION COEFFICIENT (cm²) 106 105. 104 103. 10² 10 0.2 3 GaP 2 Si 0.6 PHOTON ENERGY (CV) 1.5 GaAs InP Figure 1 In0.53 Ga0.47As 1.4 WAVELENGTH (μm) 0.7 1.8
An LED emits light with a wavelength of 500 nm (500x10-9m). What is the band gap energy in electron Volts? Hint: recall Eq. (2) and 1 eV= 1.6 x10-19 J 1.2 eV 2.5 eV 3.3 eV 4.1 eV 5.5 eV
4a.2. Using the table of experimental values of carrier concentrations n for various metals, determine the theoretical Hall coefficient RH and compare this with the experimental results provided. Calculate the experimental conductivity from the experimental results provided. Metal Silver Aluminum Gold Copper Gallium Indium Magnesium Sodium Electron concentration x 1028 m³ 5.85 18.06 5.90 8.45 15.3 11.49 8.60 2.56 Hall coefficient RH x 10-11 m³A-¹S-1 - 9.0 -3.5 -7.2 - 5.5 - 6.3 -2.4 - 9.4 - 25 Hall mobility μH = |ORH| x 10-4 m²V-1g-1 57 13 31 32 3.6 2.9 22 53
What fraction of the electrons in a good conductor have energies between 0.90 EF and EF at T = 0?
When a p-n junction is reverse biased, a) potential barrier decreases b) potential barrier increases c) majority-carrier current reduces to zero d) minority-carrier current reduces to zero
Electrons in conductors can continue to accelerate and rapidly increase indefinitely correct Error If the propagation current is opposite to the drift current, then the resultant current is the sum of the twosteams correct O wrong When the electrons reach the middle of the metal, you can never leave it for any external influence as a result The presence of a voltagebarrier correct O wrong
30
When the forward bias voltage of a diode is changed form 0.6V to 0.7V, the current changes form 5 mA to 15 mA. Then its forward bias resistance is A. 0.01 N B 10 0 Qo.1A

SEE MORE QUESTIONS