Functions, Data Structures, and File Input or Output

Pyhton 3

code

Jupyter Notebook

Numpy

Pandas

Chemistry

Physics

Author

Dinda Gusti Ayu

Published

April 6, 2025

Function, Data Structures, and File Input or Output

This work aims to applying the concept of Functions, Data Structures, and File Input/Output for solving physics problems by Python 3 and Jupyter Notebook.
There are some solutions for tasks in this Python code.

Key Equation

The Solar mass
- \(M_{Sun}=\frac{4\pi^{2}(1 AU)^{3}}{G (1 yr)^{2}}\)
The velocity of an atom
- \(v(x)=\sqrt{v_{0}^{2}+\frac{2F_{0}}{m}(cos\frac{x}{n}-1}\)

1. Function

P4.1 - The solar mass

Calculate the solar mass. The solar mass can be calculated by using the relation:

\(M_{Sun}=\frac{4\pi^{2}(1 AU)^{3}}{G (1 yr)^{2}}\)

In this task we will use approximate values for AU and G. The unit AU is an astronomical unit of length. Its value is defined to be the average distance between the Sun and Earth:

1 AU = 1.496 x \(10^{11}\) m

where 1 lightyear = 9.5 x \(10^{12}\) km. The constant G is called the gravitational constant and has the following value:

G = 6.674 X \(10^{-11}m^{3}kg^{-1}s^{-1}\)

# The Astronomical unit of length, in meters
AU = 1.496e11

# The orbital period o Earth in seconds, for one year
yr = 365.256 * 24 * 3600

# The Gravitational constant, in m3.kg-1,s-2
G = 6.674e-11

# Calculate the solar mass, in kg
import math
def Solar_mass():
    M_Sun = (4 * math.pi ** 2 * AU**3) / (G * yr**2)
    return M_Sun

print(Solar_mass())

1.9885939441528014e+30

P4.2 - The velocity of an atom

The atoms within a material is structured such that they create a lattice. We will look at an atom which moves along the surface of a material. Since the atoms are aligned as a lattice, we could use a periodic model to find the velocity of the atom moving across the surface:

\(v(x)=\sqrt{v_{0}^{2}+\frac{2F_{0}}{m}(cos\frac{x}{n}-1)}\)

where m is the mass of the atom, x is the position, v0 is its initial velocity and n is a scaled distance between the atoms within the material. We set the force F0 = 1N. Find the velocity of the atom when x = 1, v0 = 2, n = 4 and m = 3.

Repeat the calculation for Zn, Cu, and Ni for F0 = 1 x \(10^{-25}\) N

v0 = 2 # the velocity of atom
x =1 # the position of atom
m = 3 # the mass of atom
n = 4 # scaled distanced between atom within the material
F0 = 1 # The force in N

# Calculate the velocity of atom
import numpy as np
def velocity_atom():
    v_a = np.sqrt(v0**2 + (2 *F0 / m) * (np.cos(x/n)-1))
    return v_a
print(velocity_atom())

1.9948120081368812

v0 = 2 # the velocity of atom
x =1 # the position of atom
n = 4 # scaled distanced between atom within the material
F0 = 1e-25 # The force in N

# Calculate for atom Zn, Cu, and Ni
A_units = 1.6605e-27 # Atomic per units in kg

# Calculate mass of Zn, Cu, and Ni
masses = {
    "Zn": 65.38 * 1.6605e-27,
    "Cu": 63.55 * 1.6605e-27,
    "Ni": 58.69 * 1.6605e-27
}

import numpy as np
def velocity_atom(atom):
    m = masses[atom]
    velocity = v0 **2 + (2 *F0 / m) * (np.cos(x/n)-1)
    return np.sqrt(velocity) if velocity >= 0 else None

for atom in masses:
    print(f"{atom}: {velocity_atom(atom)}")

Zn: 1.9856306859195652
Cu: 1.985215363307596
Ni: 1.9839861380776256

2. Data Structures

P4.3 - The crystall lattice structures

The crystal lattice structures of the first row transition metal elements are given below. Some elements have different crystal structures under different conditions of temperatures and pressure. Use python sets to group them and determine which metal, - (a) only exist face-centered cubic (fcc), body-centered cubic (bcc), hexagonal-centered cubic (hcp) - (b) exist in two of these structures - (c) do not form an hcp structures.

Lattice	Elements
FCC	Cu, Co, Fe, Mn, Ni, Sc
BCC	Cr, Fe, Mn, Ti, V
HCP	Co, Ni, Sc, Ti, Zn

# set the elements in fcc, bcc, and hcp by lattice type
fcc = set(['Cu', 'Co', 'Fe', 'Mn', 'Ni', 'Sc'])
bcc = set(['Cr', 'Fe', 'Mn', 'Ti', 'V'])
hcp = set(['Co', 'Ni', 'Sc', 'Ti', 'Zn'])

# (a) print only exist fcc, bcc, and hcp
print('(a) only_fcc:', fcc - bcc - hcp)
print('(a) only_bcc:', bcc - fcc - hcp)
print('(a) only_hcp:', hcp - fcc - bcc)

# (b) exist in two structures
print('(b) fcc and bcc:', fcc & bcc - hcp)
print('(b) fcc and hcp:', fcc & hcp - bcc)
print('(b) bcc and hcp:', bcc & hcp - fcc)

# (c) print do not form hcp
all_metals = fcc | bcc | hcp

print('(c) no_hcp:', all_metals - hcp)

# (d) print the metals exist in fcc, bcc, and hcp
print('(d) fcc and bcc and hcp:', fcc & bcc & hcp)

(a) only_fcc: {'Cu'}
(a) only_bcc: {'V', 'Cr'}
(a) only_hcp: {'Zn'}
(b) fcc and bcc: {'Mn', 'Fe'}
(b) fcc and hcp: {'Co', 'Ni', 'Sc'}
(b) bcc and hcp: {'Ti'}
(c) no_hcp: {'V', 'Cu', 'Cr', 'Fe', 'Mn'}
(d) fcc and bcc and hcp: set()

P4.4 - Magnetic classification

The magnetic properties is related to metals. Some elements have different properties under different conditions of temperatures and pressure. Use python sets to group them and determine which category, - (a) only exist diamagnetic, paramagnetic, and non-magnetic. - (b) exist in two of these magnetic. - (c) exist in three of these magnetic.

category	Elements
Diamagnetic	Zn, Cu, Sc, Ti
Paramagnetic	Cr, Mn, V, Fe, Cu, Ti
Non-magnetic	Al, Zn, Sc

# set the elements in diamagnetic, paramagnetic and non-magnetic
Diamagnetic = set(['Zn', 'Cu', 'Sc', 'Ti'])
Paramagnetic = set(['Cr', 'Mn', 'V', 'Fe', 'Cu', 'Ti'])
Non_magnetic = set(['Al', 'Zn', 'Sc'])

# print only exist diamagnetic, paramagnetic or non-magnetic
print('(a) only_Diamagnetic:', Diamagnetic - Paramagnetic - Non_magnetic)
print('(a) only_Paramagnetic:', Paramagnetic - Diamagnetic - Non_magnetic)
print('(a) only_Nonmagnetoc:', Non_magnetic - Diamagnetic - Paramagnetic)

# Exist in two properties
print('(b) Diamagnetic and Paramagnetic:', Diamagnetic & Paramagnetic - Non_magnetic)
print('(b) Diamagnetic and Non_magnetic:', Diamagnetic & Non_magnetic - Paramagnetic)
print('(b) Paramagnetic and Non_magnetic:', Paramagnetic & Non_magnetic - Diamagnetic)

# Exist in three properties
all_category = Diamagnetic | Paramagnetic | Non_magnetic

print('(c) Diamagnetic and Paramagnetic and Non_magnetic:', Diamagnetic & Paramagnetic & Non_magnetic)

(a) only_Diamagnetic: set()
(a) only_Paramagnetic: {'V', 'Cr', 'Mn', 'Fe'}
(a) only_Nonmagnetoc: {'Al'}
(b) Diamagnetic and Paramagnetic: {'Ti', 'Cu'}
(b) Diamagnetic and Non_magnetic: {'Zn', 'Sc'}
(b) Paramagnetic and Non_magnetic: set()
(c) Diamagnetic and Paramagnetic and Non_magnetic: set()

3. File Input/Output

Table. Python file open modes

Character	Description
‘r’	open for reading (the default)
‘w’	open for writing, over writing the file if it exist
‘x’	open for writing, failing if the file exist already
‘a’	open for writing, appending if the file exist
‘t’	read/write text data (the dafault)
‘b’	read/write binary data
‘+’	open for updating (reading and writing)

P4.5 - The computational density

The text file atomic-data.txt, which can download directly, contains data for the elements, molar cell, and volume cell for Al, Cu, Pb. Read in data and calculate computational density for each subtances.

import pandas as pd

# Read the txt file (comma-separated)
df = pd.read_csv("atomic-data.txt")

print(df.head())   # show first 5 rows

for row in data:
    element, molar_cell, volume_cell = row
    molar_cell = float(molar_cell)
    volume_cell = float(volume_cell)
    density = molar_cell / volume_cell
    print(f"{element}: Density = {density:.2f} g/cm³")

  elements      \tm-cell  \tvolume-cell (cm^3)
0       Al  1.792000e-22          6.620000e-23
1       Cu  4.221000e-22          6.750000e-23
2       Pb  1.376000e-21          1.320000e-22
Al: Density = 2.71 g/cm³
Cu: Density = 6.25 g/cm³
Pb: Density = 10.42 g/cm³