To solve the given problem, we will proceed step by step.
### Given Data:
- Surface tension of liquid helium, \( \gamma = 0.35 \) mN/m = \( 0.35 \times 10^{-3} \) N/m
- Density of liquid helium, \( \rho = 145 \) kg/m³
- Mass of a helium atom, \( m_{\text{He}} = 6.64 \times 10^{-27} \) kg
### Part (a): Estimating the number of atoms per square meter in the surface
1. **Finding the number density of helium atoms in the liquid**
The number of helium atoms per unit volume (number density) is given by:
\[
n = \frac{\rho}{m_{\text{He}}}
\]
Substituting the values:
\[
n = \frac{145}{6.64 \times 10^{-27}}
\]
\[
n = 2.184 \times 10^{28} \text{ atoms/m}^3
\]
2. **Finding the atomic spacing**
We assume each atom occupies a cubic volume, so the volume per atom is:
\[
V_{\text{atom}} = \frac{1}{n}
\]
\[
V_{\text{atom}} = \frac{1}{2.184 \times 10^{28}}
\]
\[
V_{\text{atom}} = 4.58 \times 10^{-29} \text{ m}^3
\]
The cube root gives the atomic spacing \( d \):
\[
d = (V_{\text{atom}})^{1/3} = (4.58 \times 10^{-29})^{1/3}
\]
\[
d \approx 3.57 \times 10^{-10} \text{ m}
\]
3. **Estimating the number of atoms per unit area at the surface**
Since each atom occupies an area of \( d^2 \) on the surface:
\[
n_s = \frac{1}{d^2}
\]
\[
n_s = \frac{1}{(3.57 \times 10^{-10})^2}
\]
\[
n_s \approx 7.85 \times 10^{18} \text{ atoms/m}^2
\]
### Part (b): Estimating the energy per bond
1. **Energy required to remove one atom from the surface**
The energy required to create a new surface per unit area is given by the surface tension:
\[
E_{\text{surf}} = \frac{\gamma}{n_s}
\]
\[
E_{\text{surf}} = \frac{0.35 \times 10^{-3}}{7.85 \times 10^{18}}
\]
\[
E_{\text{surf}} = 4.46 \times 10^{-23} \text{ J}
\]
2. **Energy per bond**
Each atom interacts with 4 nearest neighbors, and breaking one bond contributes to half of the total energy required to remove the atom. Hence, the energy per bond is:
\[
E_{\text{bond}} = \frac{E_{\text{surf}}}{2}
\]
\[
E_{\text{bond}} = \frac{4.46 \times 10^{-23}}{2}
\]
\[
E_{\text{bond}} = 2.23 \times 10^{-23} \text{ J}
\]
3. **Converting to eV**
Since \( 1 \) eV = \( 1.6 \times 10^{-19} \) J,
\[
E_{\text{bond}} = \frac{2.23 \times 10^{-23}}{1.6 \times 10^{-19}}
\]
\[
E_{\text{bond}} \approx 1.39 \times 10^{-4} \text{ eV}
\]
### Final Answers:
(a) The number of atoms per square meter at the surface is **\( 7.85 \times 10^{18} \) atoms/m²**.
(b) The energy per bond is **\( 1.39 \times 10^{-4} \) eV**.
