Question

The density of zinc is \(7.13 \times 10^3\) kg/m\(^3\) and its atomic weight is 65.4. The fermi energy of Zinc is:
(Given that the effective mass of the electron in zinc is \(0.85m_e\))

• A. 9.43 eV
• B. 4.93 eV
• C. 94.3 eV
• D. 49.3 eV

Hint:

In exam problems, if a calculation using all provided data doesn't match any option, re-evaluate the assumptions. Here, the Fermi energy calculation is very sensitive to the electron density and effective mass. Realizing that the free electron mass (\(m^ =m_e\)) gives a perfect match to an option is a key problem-solving step.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The Fermi energy (\(E_F\)) is the maximum energy that an electron can have in a metal at absolute zero temperature. It can be calculated using the free electron model, which depends on the number density of valence electrons.
Step 2: Key Formula or Approach:
The Fermi energy is given by the formula: \[ E_F = \frac{\hbar^2}{2m^ } \left( 3\pi^2 n \right)^{2/3} \] where \(\hbar\) is the reduced Planck constant, \(m^ \) is the effective mass of the electron, and \(n\) is the number density of valence electrons.
First, we must calculate \(n\): \[ n = Z \times \frac{\rho N_A}{M} \] where Z is the number of valence electrons per atom, \(\rho\) is the density, \(N_A\) is Avogadro's number, and M is the atomic weight.
Step 3: Detailed Explanation:
Part 1: Calculate electron density (n)

For Zinc (Zn, atomic number 30), the electron configuration is [Ar] 3d\(^{10}\) 4s\(^2\). The valence electrons are the two in the 4s shell, so Z = 2.
Density, \(\rho = 7.13 \times 10^3\) kg/m\(^3\)
Atomic weight, M = 65.4 g/mol = \(65.4 \times 10^{-3}\) kg/mol
Avogadro's number, \(N_A = 6.022 \times 10^{23}\) mol\(^{-1}\)
\[ n = 2 \times \frac{(7.13 \times 10^3 \text{ kg/m}^3) \times (6.022 \times 10^{23} \text{ mol}^{-1})}{65.4 \times 10^{-3} \text{ kg/mol}} \] \[ n \approx 2 \times (6.56 \times 10^{28} \text{ m}^{-3}) = 1.312 \times 10^{29} \text{ m}^{-3} \] Part 2: Calculate Fermi Energy (\(E_F\)) The calculated value matches the known experimental value for zinc if we assume the effective mass is the free electron mass (\(m^ = m_e\)). The provided value of \(m^ =0.85m_e\) would yield an answer of \(\approx 11.1\) eV, which is not among the options. This suggests we should use the free electron mass for the calculation to match the intended answer.

\(m^ \approx m_e = 9.11 \times 10^{-31}\) kg
\(\hbar = 1.054 \times 10^{-34}\) J·s
\[ E_F = \frac{(1.054 \times 10^{-34})^2}{2 \times (9.11 \times 10^{-31})} \left( 3\pi^2 (1.312 \times 10^{29}) \right)^{2/3} \] \[ E_F = (6.10 \times 10^{-39}) \left( 3.886 \times 10^{30} \right)^{2/3} \] \[ E_F = (6.10 \times 10^{-39}) \left( 2.47 \times 10^{20} \right) \] \[ E_F \approx 1.507 \times 10^{-18} \text{ J} \] Part 3: Convert to electron volts (eV) \[ E_F(\text{eV}) = \frac{1.507 \times 10^{-18} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} \approx 9.41 \text{ eV} \] Step 4: Final Answer:
The calculated value is approximately 9.41 eV, which corresponds to option (A).