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

• A. 57 $ \overset{\text{o}}{\mathop{\text{A}}}\, $
• B. $ 57\times {{10}^{-3}} $ $ \overset{\text{o}}{\mathop{\text{A}}}\, $
• C. 217100 $ \overset{\text{o}}{\mathop{\text{A}}}\, $
• D. $ 11.61\times {{10}^{-33}} $ $ \overset{\text{o}}{\mathop{\text{A}}}\, $ (Plancks constant $ h=6.6\times {{10}^{-34}} $ Js)

Answer 1

Answer: (C)

The energy of photon $ =\frac{hc}{\lambda }, $ which should be equal to 57 MeV $ =57\times {{10}^{-3}}\times 1.6\times {{10}^{-19}}J $ Therefore, $ \lambda =\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{57\times {{10}^{-3}}\times 1.6\times {{10}^{-19}}} $ $ =0.2171\times {{10}^{-4}}m $ $ =217100\,\overset{\text{o}}{\mathop{\text{A}}}\, $