Pure silicon (density 2.33 g cm^-3, atomic mass 28) is doped with boron so that there is one boron atom per 10^7 silicon atoms. Using N_A = 6.022x10^23 mol^-1, find the number of holes created per cm^3 in units of 10^15 (two decimals), assuming each boron atom contributes one hole.

JEE Main levelNumerical answerDifficulty 2 of 3ss210

Pure silicon (density $2.33 g c m^{- 3}$ , atomic mass $28$ ) is doped with boron so that there is one boron atom per $1 0^{7}$ silicon atoms. Using $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ , find the number of holes created per $c m^{3}$ in units of $1 0^{15}$ (two decimals), assuming each boron atom contributes one hole.

Numerical answer type, answer in x10^15 per cm^3, tolerance 0.01.

Si atoms per

c m^{3} = \frac{2.33}{28} \times 6.022 \times 1 0^{23} = 5.0112 e + 22

. Boron atoms

= 5.0112 e + 22 \times 1 0^{- 7} = 5.011 \times 1 0^{15}

; each gives one hole, so holes

\approx 5.01 \times 1 0^{15} c m^{- 3}