Suppose a pure Si crystal has <math> <mn>5</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>28</mn> </mrow> </mrow> </msup> </math> atoms <math> <mo>/</mo> <msup> <mrow> <mrow> <mi mathvariant="normal">m</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </math> . It is doped by <math> <mn>1</mn> <mi mathvariant="normal">p</mi> <mi mathvariant="normal">p</mi> <mi mathvariant="normal">m</mi> </math> concentration of pentavalent As. The number of holes <math> <mfenced separators="|"> <mrow> <mrow> <msub> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mi>i</mi> </mrow> </mrow> </msub> <mo>=</mo> <mn>1.5</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>16</mn> </mrow> </mrow> </msup> <mo>/</mo> <msup> <mrow> <mrow> <mi mathvariant="normal">m</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfenced> </math> is:

n i 2 = n e × n h where n e = 5 × 10 28 10 6 ∴ n h = n i 2 n e = 1.5 × 10 16 1.5 × 10 16 × 10 6 5 × 10 28 n h = 4.5 × 10 9 / m 3

&nbsp;Suppose a pure Si crystal has <math> <mn>5</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>28</mn> </mrow> </mrow> </msup> </math> atoms <math> <mo>/</mo> <msup> <mrow> <mrow> <mi mathvariant="normal">m</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </math> . It is doped by <math> <mn>1</mn> <mi mathvariant="normal">p</mi> <mi mathvariant="normal">p</mi> <mi mathvariant="normal">m</mi> </math> concentration of pentavalent As. The number of holes <math> <mfenced separators="|"> <mrow> <mrow> <msub> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mi>i</mi> </mrow> </mrow> </msub> <mo>=</mo> <mn>1.5</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>16</mn> </mrow> </mrow> </msup> <mo>/</mo> <msup> <mrow> <mrow> <mi mathvariant="normal">m</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfenced> </math> &nbsp;is: &lt;!-- [if !supportLineBreakNewLine]--&gt; &lt;!--[endif]--&gt;

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Physics Class 12th Physics Semiconductor Devices

Suppose a pure Si crystal has $5 \times 10^{28}$ atoms $/ m^{3}$ . It is doped by $1 p p m$ concentration of pentavalent As. The number of holes $(n_{i} = 1.5 \times 10^{16} / m^{3})$ is:

Option 1 -

$4.5 * 10^{9} / m^{3}$

Option 2 -

$7 * 10^{9} / m^{3}$

Option 3 -

$1.5 * 10^{12} / m^{3}$

Option 4 -

$4 * 10^{8} / m^{3}$

2 Views | Posted 3 weeks ago

Asked by Shiksha User

1 Answer

Answered by

Raj Pandey | Contributor-Level 9

3 weeks ago

Correct Option - 1

Detailed Solution:

${(n_{i})}^{2} = n_{e} \times n_{h}$ where $n_{e} = \frac{5 \times 10^{28}}{10^{6}}$

$\begin{array}{r} ∴ n_{h} & = \frac{{(n_{i})}^{2}}{n_{e}} = \frac{(1.5 \times 10^{16}) (1.5 \times 10^{16}) \times 10^{6}}{5 \times 10^{28}} \\ n_{h} & = 4.5 \times 10^{9} / m^{3} \end{array}$

<math> <msup> <mrow> <mrow> <mfenced separators="|"> <mrow> <mrow> <msub> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mi>i</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>=</mo> <msub> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> </msub> <mo>×</mo> <msub> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> </msub> </math> where <math> <msub> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>5</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>28</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> <math> <mtable columnalign="right"> <mtr> <mtd> <mrow> <mo>∴</mo> <mrow> <mi mathvariant="normal"> </mi> </mrow> <msub> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> </msub> </mrow> </mtd> <mtd> <mrow> <mi> </mi> <mo>=</mo> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mfenced separators="|"> <mrow> <mrow> <msub> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mi>i</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mfenced separators="|"> <mrow> <mrow> <mn>1.5</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>16</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfenced> <mfenced separators="|"> <mrow> <mrow> <mn>1.5</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>16</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfenced> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>28</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> </msub> </mrow> </mtd> <mtd> <mrow> <mi> </mi> <mo>=</mo> <mn>4.5</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>9</mn> </mrow> </mrow> </msup> <mo>/</mo> <msup> <mrow> <mrow> <mi mathvariant="normal">m</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </mrow> </mtd> </mtr> </mtable> </math>

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Suppose a pure Si crystal has $5 \times 10^{28}$ atoms $/ m^{3}$ . It is doped by $1 p p m$ concentration of pentavalent As. The number of holes $(n_{i} = 1.5 \times 10^{16} / m^{3})$ is:

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Suppose a pure Si crystal has 5 × 10 28 atoms / m 3 . It is doped by 1 p p m concentration of pentavalent As. The number of holes n i = 1.5 × 10 16 / m 3 is:

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Suppose a pure Si crystal has $5 \times 10^{28}$ atoms $/ m^{3}$ . It is doped by $1 p p m$ concentration of pentavalent As. The number of holes $(n_{i} = 1.5 \times 10^{16} / m^{3})$ is: