Question

2. How far below (or above) the Fermi energy can states be found which are 90% occupied by electrons at 300 K? Repeat this calculation for 99%, 10% and 1% Consider what would happen if instead of using periodic boundary conditions, the following the boundary conditions b. 40, y, z, t) 0, (L,y, z,t) 0, ?(z, 0, z, t)-0, y(z, L, z, t) 0, v(x,g, 0, t) 0, are retained for particles in a box. Find the resulting density of states in k-space. Show that the number of states N(E) with energy less than some value, E, is given by:

Please Write the answer very clear By using PC.

Answer 1

Probabilty of occupancy of an energy level E below E_f is:

$^{f_{(E)}}=\frac{1}{1+exp(-\Delta E/kT)}$

we have k = 8.62 * 10^-5 eVK^-1 and T = 300 K,

kT = 0.02586 eV,

For f_(E) = 90% = 0.90,

$0.90=\frac{1}{1+exp(-\Delta E/0.02586)}$

Making Reciprocal and taking natural log on both side,

$ln(1/0.90)=ln (exp(\frac{-\Delta E}{0.02586}))$

by Solving,

$\Delta E = -0.0027246 eV$ (Above Fermi Energy Level)  

For f_(E) = 10% = 0.10,

$0.10=\frac{1}{1+exp(-\Delta E/0.02586)}$

Making Reciprocal and taking natural log on both side,

$ln(1/0.10)=ln (exp(\frac{-\Delta E}{0.02586}))$

by Solving,

$\Delta E = -0.05954 eV$ (Above Fermi Energy Level)  

For f_(E) = 1% = 0.001,

$0.001=\frac{1}{1+exp(-\Delta E/0.02586)}$

Making Reciprocal and taking natural log on both side,

$ln(1/0.001)=ln (exp(\frac{-\Delta E}{0.02586}))$

by Solving,

$\Delta E = -0.1190 eV$ (Above Fermi Energy Level)