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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# What is the probability that, at a temperature of T = 300 K, an electron will jump across the energy gap ${{\mathbf{E}}}_{{\mathbf{g}}}\left(=5.5\mathrm{eV}\right)$ in a diamond that has a mass equal to the mass of Earth? Use the molar mass of carbon in Appendix F; assume that in diamond there is one valence electron per carbon atom.

The probability that an electron will jump across the energy gap in a diamond is $1.28×{10}^{-42}$

See the step by step solution

## Step 1: The given data

a) Temperature of the energy state, T = 300 K

b) Value of energy gap, ${E}_{g}=5.5\mathrm{eV}$

c) Mass of Earth, ${M}_{e}=5.98×{10}^{24}\mathrm{kg}$

d) Molar mass of carbon, m = 12.01 g/ mol

e) Assumption that in diamond there is one valence electron per carbon atom

## Step 2: Understanding the concept of Fermi energy

The highest energy level occupied by an electron in the valence band at absolute zero temperature is known as Fermi level and the energy of electrons present in that level is known as Fermi energy.

Formulae:

Number of excited atoms in a mass,

${N}_{e}=\frac{{M}_{e}}{m}{N}_{A}$ (i)

$\mathrm{where}{\mathrm{N}}_{\mathrm{A}}=6.02×{10}^{23}$ is called Avogadro’s number.

The probability of an electron to get excited in an insulator,

$P={N}_{e}^{-{E}_{g}/kT}$ (ii)

$\mathrm{where}\mathrm{k}=8.62×{10}^{-5}\mathrm{eV}$

## Step 3: Calculation of the probability of excitation atoms to jump an given energy band gap

The number of carbon atoms in a diamond, as massive as the Earth, is given by the number of electrons that get excited.

Thus, using equation (i) and equation (ii), the probability for an electron to get excited is given as-

$P=\left(\frac{{M}_{e}}{m}{N}_{A}\right){e}^{\left(-{E}_{g}/kT\right)}\phantom{\rule{0ex}{0ex}}=\left(\frac{\left(5.98×{10}^{24}\mathrm{kg}\right)\left(6.02×{10}^{23}/\mathrm{mol}\right)}{12.01\mathrm{g}/\mathrm{mol}}\right){e}^{\left(\frac{-5.5\mathrm{eV}}{\left(8.62×{10}^{-5}\mathrm{eV}/\mathrm{K}\right)300\mathrm{K}}\right)}\phantom{\rule{0ex}{0ex}}=1.28×{10}^{-42}$

Hence, the value of the probability is $1.28×{10}^{-42}$.