Electronic Devices Question 230
Question: A P-type semiconductor has acceptor levels 57 meV above the valence band. The maximum wavelength of light required to create a hole is (Planck’s constant h = $ 6.6\times {{10}^{-34}} $ J-s)
[MP PET 1995]
Options:
A) $ 57{AA} $
B) $ 57\times {{10}^{-3}}{AA} $
C) $ 217100{AA} $
D) $ 11.61\times {{10}^{-33}}{AA} $
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Answer:
Correct Answer: C
Solution:
$ E=\frac{hc}{\lambda } $
Therefore $ \lambda =\frac{hc}{E} $
$ =\frac{6.6\times {{10}^{-34}}\times 3\times 10^{8}}{57\times {{10}^{-3}}\times 1.6\times {{10}^{-19}}} $
$ =217100{AA} $ .
