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A silicon bar is doped with 1017 arsenic atoms /cm3. The equilibrium hole concentration at 300k and the location of the Fermi level (EF) of sample relative to intrinsic Fermi level (EFi) are(Given, ni = 1.5 × 1010/cm3 )a)2.25×103 /cm3and 0.4 eV up respectivelyb)2.25×104 /cm3and 0.6 eV up respectivelyc)2.25 ×103 /cm3 and 0.4 eV down respectivelyd)2.25 ×104 /cm3 and 0.4 eV down respectivelyCorrect answer is option 'A'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared
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the Electronics and Communication Engineering (ECE) exam syllabus. Information about A silicon bar is doped with 1017 arsenic atoms /cm3. The equilibrium hole concentration at 300k and the location of the Fermi level (EF) of sample relative to intrinsic Fermi level (EFi) are(Given, ni = 1.5 × 1010/cm3 )a)2.25×103 /cm3and 0.4 eV up respectivelyb)2.25×104 /cm3and 0.6 eV up respectivelyc)2.25 ×103 /cm3 and 0.4 eV down respectivelyd)2.25 ×104 /cm3 and 0.4 eV down respectivelyCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam.
Find important definitions, questions, meanings, examples, exercises and tests below for A silicon bar is doped with 1017 arsenic atoms /cm3. The equilibrium hole concentration at 300k and the location of the Fermi level (EF) of sample relative to intrinsic Fermi level (EFi) are(Given, ni = 1.5 × 1010/cm3 )a)2.25×103 /cm3and 0.4 eV up respectivelyb)2.25×104 /cm3and 0.6 eV up respectivelyc)2.25 ×103 /cm3 and 0.4 eV down respectivelyd)2.25 ×104 /cm3 and 0.4 eV down respectivelyCorrect answer is option 'A'. Can you explain this answer?.
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Here you can find the meaning of A silicon bar is doped with 1017 arsenic atoms /cm3. The equilibrium hole concentration at 300k and the location of the Fermi level (EF) of sample relative to intrinsic Fermi level (EFi) are(Given, ni = 1.5 × 1010/cm3 )a)2.25×103 /cm3and 0.4 eV up respectivelyb)2.25×104 /cm3and 0.6 eV up respectivelyc)2.25 ×103 /cm3 and 0.4 eV down respectivelyd)2.25 ×104 /cm3 and 0.4 eV down respectivelyCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A silicon bar is doped with 1017 arsenic atoms /cm3. The equilibrium hole concentration at 300k and the location of the Fermi level (EF) of sample relative to intrinsic Fermi level (EFi) are(Given, ni = 1.5 × 1010/cm3 )a)2.25×103 /cm3and 0.4 eV up respectivelyb)2.25×104 /cm3and 0.6 eV up respectivelyc)2.25 ×103 /cm3 and 0.4 eV down respectivelyd)2.25 ×104 /cm3 and 0.4 eV down respectivelyCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A silicon bar is doped with 1017 arsenic atoms /cm3. The equilibrium hole concentration at 300k and the location of the Fermi level (EF) of sample relative to intrinsic Fermi level (EFi) are(Given, ni = 1.5 × 1010/cm3 )a)2.25×103 /cm3and 0.4 eV up respectivelyb)2.25×104 /cm3and 0.6 eV up respectivelyc)2.25 ×103 /cm3 and 0.4 eV down respectivelyd)2.25 ×104 /cm3 and 0.4 eV down respectivelyCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A silicon bar is doped with 1017 arsenic atoms /cm3. The equilibrium hole concentration at 300k and the location of the Fermi level (EF) of sample relative to intrinsic Fermi level (EFi) are(Given, ni = 1.5 × 1010/cm3 )a)2.25×103 /cm3and 0.4 eV up respectivelyb)2.25×104 /cm3and 0.6 eV up respectivelyc)2.25 ×103 /cm3 and 0.4 eV down respectivelyd)2.25 ×104 /cm3 and 0.4 eV down respectivelyCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A silicon bar is doped with 1017 arsenic atoms /cm3. The equilibrium hole concentration at 300k and the location of the Fermi level (EF) of sample relative to intrinsic Fermi level (EFi) are(Given, ni = 1.5 × 1010/cm3 )a)2.25×103 /cm3and 0.4 eV up respectivelyb)2.25×104 /cm3and 0.6 eV up respectivelyc)2.25 ×103 /cm3 and 0.4 eV down respectivelyd)2.25 ×104 /cm3 and 0.4 eV down respectivelyCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Electronics and Communication Engineering (ECE) tests.