Magnetic Effects Of Current And Magnetism

Step 1: Understanding Molecular Orbital Theory.
According to Molecular Orbital (MO) theory, paramagnetism occurs when a molecule has one or more unpaired electrons in its molecular orbitals. These unpaired electrons create a magnetic field, making the molecule attracted to a magnetic field (paramagnetic behavior). Diamagnetic molecules, on the other hand, have all their electrons paired and do not exhibit such attraction.
Step 2: Molecular Orbitals for Diatomic Molecules.
Let’s apply the molecular orbital theory to the given molecules to check for unpaired electrons.
1. Oxygen ({ O }₂):
The electronic configuration for { O }₂ (in the molecular orbital theory) is:

 ( \sigma ₁s², \sigma ^{₁s}², \sigma ₂s², \sigma ^{₂s}², \pi ₂p̀x;², \pi ₂p̀y;², \pi ^{₂p̀x;}¹, \pi ^{₂p̀y;}¹ ) 

Here, the two \pi -antibonding orbitals each have one unpaired electron, making { O }₂ paramagnetic.
2. Nitrogen ({ N }₂):
The electronic configuration for { N }₂ is:

 ( \sigma ₁s², \sigma ^{₁s}², \sigma ₂s², \sigma ^{₂s}², \pi ₂p̀x;², \pi ₂p̀y;² ) 

All electrons in { N }₂ are paired, so it is diamagnetic and does not exhibit paramagnetism.
3. Fluorine ({ F }₂):
The electronic configuration for { F }₂ is:

 ( \sigma ₁s², \sigma ^{₁s}², \sigma ₂s², \sigma ^{₂s}², \pi ₂p̀x;², \pi ₂p̀y;², \pi ^{₂p̀x;}², \pi ^{₂p̀y;}² ) 

All electrons are paired in { F }₂, so it is diamagnetic and does not exhibit paramagnetism.
4. Carbon ({ C }₂):
The electronic configuration for { C }₂ is:

 ( \sigma ₁s², \sigma ^{₁s}², \sigma ₂s², \sigma ^{₂s}², \pi ₂p̀x;², \pi ₂p̀y;² ) 

All electrons are paired in { C }₂, so it is diamagnetic and does not exhibit paramagnetism.
Step 3: Conclusion.
Based on the molecular orbital theory, { O }₂ has unpaired electrons and exhibits paramagnetism. The correct answer is (1) { O }₂.

Step 1: Understanding the formula for the radius of a charged particle in a magnetic field.
The radius of the circular path of a charged particle moving perpendicular to a magnetic field is given by the formula:

Where:
is the mass of the particle,
is the velocity of the particle,
is the charge of the particle,
is the magnetic field strength.
Step 2: Relation between kinetic energy and velocity.
The kinetic energy of a particle is given by:

For both the proton and the deuteron, the kinetic energy is equal, so we can express the velocity in terms of kinetic energy as:

Thus, both particles have the same kinetic energy, and the expression for the velocity will be similar for both.
Step 3: Radius of the proton and the deuteron.
Substituting for into the formula for radius:

Since both the proton and the deuteron have the same kinetic energy, their radii will depend on their masses and charges. The charge of both the proton and the deuteron is the same (both carry the charge of a proton, ), but the mass of the deuteron is twice that of the proton.
Step 4: Ratio of radii.
Let the mass of the proton be and the mass of the deuteron be . The radius of the proton and the radius of the deuteron can be written as:

Thus, the ratio of the radii is:

Therefore, the ratio of the radii of the proton's circular path to the deuteron's circular path is 1.
Step 5: Conclusion.
The correct answer is 1, which corresponds to option (1).