A) \[18.855\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
B) \[108.55\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
C) \[1085.5\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
D) \[10855\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
Correct Answer: D
Solution :
Energy absorbed by silicon is given by \[\Delta \Epsilon =\frac{hc}{\lambda }\]or \[\lambda =\frac{hc}{\Delta \Epsilon }\] Here, \[h=6.6\times {{10}^{-34}}\,J-s,\]\[c=3\times {{10}^{8}}\,m/s,\] \[\Delta \Epsilon =1.14\,eV=1.14\times 1.6\times {{10}^{-19}}J\] \[\therefore \] \[\lambda =\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{1.14\times 1.6\times {{10}^{-19}}}\] \[=10.855\times {{10}^{-7}}\] \[=10855\,\overset{\text{o}}{\mathop{\text{A}}}\,\] Note: After putting the values of and c in the expression for energy, we get \[\Delta E=\frac{12375}{\lambda (\overset{\text{o}}{\mathop{\text{A}}}\,)}eV\] or \[\lambda =\frac{12375}{\Delta E(eV)}\overset{\text{o}}{\mathop{\text{A}}}\,\]
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