What diameter does an aluminum particle (a perfect shape) achieve quantum confin

Question

What diameter does an aluminum particle (a perfect shape) achieve quantum confinement at room temperature (25 degrees Celsius)? An aluminum atom has the electronic configuration [Ne]3s^2 3p^1, giving three valence electrons. The density of solid aluminum is 2.70 g/cm^3. its atomic mass is 27 g/mol. The aluminum Fermi energy is 11.63 eV. What diameter does an aluminum particle (a perfect shape) achieve quantum confinement at room temperature (25 degrees Celsius)? An aluminum atom has the electronic configuration [Ne]3s^2 3p^1, giving three valence electrons. The density of solid aluminum is 2.70 g/cm^3. its atomic mass is 27 g/mol. The aluminum Fermi energy is 11.63 eV.

Explanation / Answer

For quantum confinement, 4Ef/3nv>kBT. Therefore
Ef=11.63 eV
So nv<(4*11.63*1.6*10^-19)/(3*298*1.38*10^-23) . So nv<606
Al has 3 valence electron. So effectively only 202 atoms are required to be confined.
So density of atoms in a given volume = 2700*6.02 * 10^23 / 0.027 = 6.02*10^28
Volume of a particle V = 202/(6.02*10^28) = 3.355*10^-27 m3
SO diameter = 2(3V/(4*pi))^.33 = 1.28*10^-9

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