The number of silicon atoms per $$m3$$ is $$5$$×$$1028$$. This is doped simultaneously with $$5$$×$$1022$$ atoms per $$m3$$ of arsenic and $$5$$×$$1020$$ atoms per $$m3$$ of Indium .Given that $$ni=$$ $$1.5$$ × $$1016$$ m−$$3$$. If the number of electrons is z×$$1021m$$−$$3$$ then z is

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The number of silicon atoms per  $$m3$$ is  $$5$$×$$1028$$. This is doped simultaneously with $$5$$×$$1022$$ atoms per $$m3$$ of arsenic and  $$5$$×$$1020$$ atoms per $$m3$$ of Indium .Given that $$ni=$$ $$1.5$$ × $$1016$$ m−$$3$$. If the number of electrons is z×$$1021m$$−$$3$$ then z is

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Donor Density (Arsenic) $$=$$ $$ND=5$$×$$1022m$$−$$3Acceptor$$ Density (Indium) $$=$$ $$NA=5$$×$$1020m$$−$$3ni=1.5$$×$$1016m$$−$$3$$ If  the semi conductor is electrically neutral , then ND−$$NA=ne$$−nh.......(1)here$$ ND−NA $$=$$  $$5$$×$$1022$$−$$5$$×$$1020=5$$×$$1020(102$$−$$1) $$=5$$×$$1020$$×$$99$$ $$=$$ $$495$$×$$1020Also$$, $$nenh=ni2$$...............(2)Now$$, we can write $$ne+nh2=(ne$$−$$nh)2+4nenh$$.....................$$(3)Put$$ $$(1) and (2) in (3) ⇒(ne+nh)=(ND$$−$$NA)2+4ni2$$⇒$$ne+nh+ne$$−$$nh=ND$$−$$NA+(ND$$−$$NA)2+4ni2$$⇒$$2ne=ND$$−$$NA+(ND$$−$$NA)2+4ni2$$⇒$$2ne=495$$×$$1020+(495$$×$$1020)2+4$$×$$(1.5$$×$$1016)2$$⇒$$2ne=495$$×$$1020+(495$$×$$495$$×$$1040)+4$$×$$(2.25$$×$$1032)⇒$$2ne$$≈$$495$$×$$1020+4952$$×$$1040$$⇒$$2ne$$≈$$495$$×$$1020+495$$×$$1020$$⇒$$2ne$$≈$$495$$×$$1020$$×$$2$$⇒ne≈$$495$$×$$1020=49.5$$×$$1021$$

The number of silicon atoms per m3 is 5×1028. This is doped simultaneously with 5×1022 atoms per m3 of arsenic and 5×1020 atoms per m3 of Indium .Given that ni= 1.5 × 1016 m−3. If the number of electrons is z×1021m−3 then z is

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