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