Determine if the plane (112) is perpendicular to the following directions:[100],

Question

Determine if the plane (112) is perpendicular to the following directions:[100], [001], [2bar11], and [1bar 1bar 0]. For cubic lattice, calculate inter-planer distances between the planes (123) and between the planes (321). Calculate angle at which these acts of planes cross each other. For Si lattice, identify the planes comprising at least three of the highlighted atoms: at what temperature density of states in conduction band of silicon and germanium will reach 5 times 10^19 cm^-3? Calculate concentration of holes in intrinsic germanium at a temperature of 100degreeC. Find position of Fermi level GaAs at a temperature of 200degreeC. Calculate position of Fermi level in silicon doped with phosphorous at a concentration of 10^11 cm^-3 and compare it with the known ionization energy of donors. Calculate of electrons in conduction band of germanium at a temperature of 100C if it is doped with phosphorous at a concentration of 10^18 cm^-3 and boron at a concentration of 3 times 10^17 cm^-3.

Explanation / Answer

Two Miller planes with miller indices : [h_{1},k_{1},l_{1}] and [h_{2},k_{2},l_{2}] are said to be perpendicular if the angle (phi) between the two planes is 90^{degree} i.e. cos{phi}=0. The angle between any two planes is obtained using the relation - cos{phi}=rac{h_{1}h_{2}+k_{1}k_{2}+l_{1}l_{2}}{sqrt{h_{1}^{2}+k_{1}^{2}+l_{1}^{2}} sqrt{h_{2}^{2}+k_{2}^{2}+l_{2}^{2}}} 1. We are given - [h_{1},k_{1},l_{1}]=[1 1 2]Rightarrow h_{1}=1,k_{1}=1,l_{1}=2\\ herefore hspace{0.2cm} sqrt{h_{1}^{2}+k_{1}^{2}+l_{1}^{2}}=sqrt{6} If [h_{2},k_{2},l_{2}]=[1 0 0]Rightarrow h_{2}=1,k_{2}=0,l_{2}=0\\ herefore hspace{0.2cm} sqrt{h_{2}^{2}+k_{2}^{2}+l_{2}^{2}}=1 therefore, cos{phi}=rac{h_{1}h_{2}+k_{1}k_{2}+l_{1}l_{2}}{sqrt{h_{1}^{2}+k_{1}^{2}+l_{1}^{2}} sqrt{h_{2}^{2}+k_{2}^{2}+l_{2}^{2}}} =rac{1+0+0}{sqrt{6}}=rac{1}{sqrt{6}} This implies, Miller plane [1 1 2] is not perpendicular to [1 0 0]. If [h_{2},k_{2},l_{2}]=[0 0 1]Rightarrow h_{2}=0,k_{2}=0,l_{2}=1\\ herefore hspace{0.2cm} sqrt{h_{2}^{2}+k_{2}^{2}+l_{2}^{2}}=1 therefore, cos{phi}=rac{h_{1}h_{2}+k_{1}k_{2}+l_{1}l_{2}}{sqrt{h_{1}^{2}+k_{1}^{2}+l_{1}^{2}} sqrt{h_{2}^{2}+k_{2}^{2}+l_{2}^{2}}} =rac{0+0+2}{sqrt{6}}=rac{2}{sqrt{6}} This implies, Miller plane [1 1 2] is not perpendicular to [0 0 1]. If [h_{2},k_{2},l_{2}]=[ar{2} 1 1]Rightarrow h_{2}=-2,k_{2}=1,l_{2}=1\\ herefore hspace{0.2cm} sqrt{h_{2}^{2}+k_{2}^{2}+l_{2}^{2}}=sqrt{6} therefore, cos{phi}=rac{h_{1}h_{2}+k_{1}k_{2}+l_{1}l_{2}}{sqrt{h_{1}^{2}+k_{1}^{2}+l_{1}^{2}} sqrt{h_{2}^{2}+k_{2}^{2}+l_{2}^{2}}} =rac{-2+1+2}{sqrt{6}sqrt{6}}= rac{1}{6} This implies, Miller plane [1 1 2] is not perpendicular to [ar{2}, extbf{1,1}]. If [h_{2},k_{2},l_{2}]=[ar{1} ar{1} 0]Rightarrow h_{2}=-1,k_{2}=-1,l_{2}=0\\ herefore hspace{0.2cm} sqrt{h_{2}^{2}+k_{2}^{2}+l_{2}^{2}}=2 therefore, cos{phi}=rac{h_{1}h_{2}+k_{1}k_{2}+l_{1}l_{2}}{sqrt{h_{1}^{2}+k_{1}^{2}+l_{1}^{2}} sqrt{h_{2}^{2}+k_{2}^{2}+l_{2}^{2}}} =rac{-1-1+0}{sqrt{6}sqrt{2}}=-rac{2}{sqrt{6}sqrt{2}} This implies, Miller plane [1 1 2] is not perpendicular to [ar{1},ar{1}, extbf{0}]. 2. The interplanar distance between two planes with indices : [h_{1},k_{1},l_{1}] and [h_{2},k_{2},l_{2}] for a cubic lattice with lattice constant (a) is written as - d_{2}-d_{1}=Delta d=aleft [ rac{1}{sqrt{h_{2}^{2}+k_{2}^{2}+l_{2}^{2}}}- rac{1}{sqrt{h_{1}^{2}+k_{1}^{2}+l_{1}^{2}}} ight ] Thus, if [h_{1},k_{1},l_{1}]=[1 2 3] and [h_{2},k_{2},l_{2}]=[3 2 1] such that sqrt{h_{1}^{2}+k_{1}^{2}+l_{1}^{2}}=sqrt{1+4+9}=sqrt{14} and sqrt{h_{2}^{2}+k_{2}^{2}+l_{2}^{2}}=sqrt{9+4+1}=sqrt{14} Thus, interplanar distance between these two planes are - Delta d=aleft [ rac{1}{sqrt{h_{2}^{2}+k_{2}^{2}+l_{2}^{2}}}- rac{1}{sqrt{h_{1}^{2}+k_{1}^{2}+l_{1}^{2}}} ight ] =aleft [ rac{1}{sqrt{14}}- rac{1}{sqrt{14}} ight ] =0 and, the angle at which these two planes cross each other is - cos{phi}=rac{h_{1}h_{2}+k_{1}k_{2}+l_{1}l_{2}}{sqrt{h_{1}^{2}+k_{1}^{2}+l_{1}^{2}} sqrt{h_{2}^{2}+k_{2}^{2}+l_{2}^{2}}} =rac{3+4+3}{sqrt{14}sqrt{14}} =rac{10}{14}=rac{5}{7} This gives us - phi=cos^{-1}{rac{5}{7}}=44.415^{degree}

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