Assuming dopant atoms are uniformly distributed in a silicon crystal, how far apart are these atoms when the doping concentrations are 5.0x1015 cm-3 and 5.0x1020 cm-3?

d =~ 5.8μm

d =~ 0.13 μm

Explanation:

when the doping concentrations are 5 × 10^15 cm^-3

d = v^-1/3  ; where d represent the distance between the atoms , and v  represent the volume

d =1/ ∛v

d = 1/ ∛5 × 10^15

d = 1/ 170997.5

d = 5.85 × 10 ^ -6

d =~ 5.8μm

when the doping concentrations are 5 × 10^20 cm^-3

d = v^-1/3  ; where d represent the distance between the atoms , and v  represent the volume

d =1/ ∛v

d = 1/ ∛5 × 10^20

using the principle of surds and standard forms, we have

d = 1/ ∛0.5 × 10^21  

d = 1/7937005.26

d = 1.26 × 10 ^ -7

d = 0.126 × 10 ^ -6

d =~ 0.13 μm

the answer is   c.) 1575 j.

explanation:

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol