Calc Electron & Hole Concen in Silicon at 300K

[bobred]

Homework Statement

For silicon at T=300K with donor density N_D=2×10⁹cm⁻³, acceptor density N_A=0 and n_i=8.2×10⁹cm⁻³, calculate the equilibrium electron and hole concentration

Homework Equations

n_0=\frac{N_D-N_A}{2}+\sqrt{\frac{N_D-N_A}{2}^2+n_i^2}

p_o=\frac{n_i^2}{n_0}

The Attempt at a Solution

I get
n_0=9.26\times10^9 and p_0=7.26\times10^9
But whenever I enter the results into the online quiz it says it's wrong, am I missing something here?

 

[NateTheGreatt77]

$$n_0= \frac{N_D-N_A}{2}+\sqrt{\frac{N_D-N_A}{2}^2+n_i^2}$$​

Check your electron concentration equation again, you may be missing a parenthesis inside the square root.

 

[bobred]

Yes there is a typo, the calculation was correct though, it should be

<br /> n_0=\frac{N_D-N_A}{2}+\sqrt{\frac{(N_D-N_A)}{2}^2+n_i^2}<br />

 

[TSny]

Homework Statement

For silicon at T=300K with donor density N_D=2×10⁹cm⁻³, acceptor density N_A=0 and n_i=8.2×10⁹cm⁻³, calculate the equilibrium electron and hole concentration

This might not be relevant, but the value of $n_i$ given here for Si at 300 K is somewhat smaller than values that I have found in a search on the internet (which tend to be between 1.0 x 10¹⁰ cm^-3 and 1.5 x 10¹⁰ cm^-3). But I did come across one site that listed a value close to your value. Certainly, if the value you used is the value that is given in the statement of the problem, then you should use it.

 

[bobred]

I have seen similar values too, but the value was given on a sheet of constants. Can't see where the problem is.

 

[TSny]

Yes there is a typo, the calculation was correct though, it should be

<br /> n_0=\frac{N_D-N_A}{2}+\sqrt{\frac{(N_D-N_A)}{2}^2+n_i^2}<br />

Still looks like a typo is present. The denominator of 2 inside the root should be squared. But I agree with your calculated value for $n_0$.

 

[bobred]

You are correct TSny, I was rushing and didn't check it properly.