This question is incomplete, the complete question is;
The number density of conduction electrons in pure silicon at room temperature is about 10¹⁶ m⁻³. By doping with phosphorous, you desire to increase this number by a factor of 1 million (10⁶). At room temperature, assume that the thermal energy is sufficient that all of the extra electrons from the P doping enter the conduction band. What fraction of silicon atoms must you replace with phosphorous atoms
Answer:
the fraction of silicon atoms we must replace with phosphorous atoms is 4( 10⁶ - 1 )
Step-by-step explanation:
Given that;
the number density no = 10¹⁶ m⁻³
Required no = (10¹⁶ × 10⁶) = 10²² m⁻³
Each silicon atom has 4 valence electron,
Hence; Silicon density = 10¹⁶/4
now, if x = No of silicon atoms to be replaced with phosphorous, then;
(10¹⁶/4 - x)4 + 5x = 10²²
10¹⁶ - 4x + 5x = 10²²
10¹⁶ + x = 10²²
x = 10²² - 10¹⁶
Fraction of silicon atoms will be;
⇒ 10²² - 10¹⁶ / 10¹⁶/4
⇒ 4( 10²² - 10¹⁶ / 10¹⁶ )
⇒ 4( 10²²/10¹⁶ - 10¹⁶/10¹⁶ )
⇒ 4( 10⁶ - 1 )
Therefore, the fraction of silicon atoms we must replace with phosphorous atoms is 4( 10⁶ - 1 )