$ 5 \times {10^7} $ સિલીકોન પરમાણુ દીઠ એક ઇન્ડિયમ પરમાણુ રાખવામાં આવે છે,સિલીકોનની પરમાણુ સંખ્યા ઘનતા $ 5 \times {10^{28}}{\rm{atoms}} /{m^3} $ હોય,તો એકસેપટર પરમાણુની સંખ્યા ઘનતા કેટલી થાય?

Question

A$ 2.5 \times {10^{30}}atoms/c{m^3} $
B$ 1.0 \times {10^{13}}atoms/c{m^3} $
C$ 1.0 \times {10^{15}}atoms/c{m^3} $
D$ 2.5 \times {10^{36}}atoms/c{m^3} $

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Solution

c
(c)Number density of atoms in silicon specimen $= 5 × 10^{28} \,\,atom/m^{3} = 5 × 10^{22}\,\, atom/cm^{3}$
Since one atom of indium is doped in $5 × 10^{7}$ Si atom. So number of indium atoms doped per $cm^3$ of silicon.$n = \frac{{5 \times {{10}^{22}}}}{{5 \times {{10}^7}}} = 1 \times {10^{15}}\,atom/c{m^3}$

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