Physics, Mathematics and Chemistry Online Academy

Bohr's Atomic Radius
Bohr's atomic radius is the average distance between the nucleus and the electron in a particular orbit of a hydrogen-like atom according to Bohr's atomic model. In 1913, Niels Bohr proposed that electrons revolve around the nucleus in fixed circular orbits without radiating energy. Each orbit has a definite radius and energy. The radius of the first orbit of the hydrogen atom is known as the Bohr radius, denoted by a₀, and it serves as the fundamental unit for measuring atomic dimensions.
The Bohr radius is given by:
a₀ = 5.29 × 10⁻¹¹ m = 0.529 Å

➡️ Formula for Bohr's Atomic Radius
The radius of the nth orbit of a hydrogen-like atom is expressed as:
rₙ = (n²a₀)/Z
Where:
rₙ = Radius of the nth orbit
n = Principal quantum number (1, 2, 3, …)
a₀ = Bohr radius (5.29 × 10⁻¹¹ m)
Z = Atomic number of the atom
For the hydrogen atom, where Z = 1, the equation becomes:
rₙ = n²a₀
This formula shows that the radius of an electron orbit increases as the square of the principal quantum number.

➡️ Derivation of Bohr's Atomic Radius
The derivation of Bohr's atomic radius is based on two important principles: Coulomb's law of electrostatic attraction and Bohr's quantization of angular momentum.

• Electrostatic Force
The electron is attracted to the positively charged nucleus by the electrostatic force, which is given by Coulomb's law:
F = (1/4πε₀) × (Ze²/r²)
where:
ε₀ = Permittivity of free space
Z = Atomic number
e = Electronic charge
r = Radius of the orbit

• Centripetal Force
As the electron moves in a circular orbit, it requires a centripetal force given by:
F = mv²/r
where:
m = Mass of the electron
v = Velocity of the electron
Since the electrostatic force provides the centripetal force,
mv²/r = (1/4πε₀) × (Ze²/r²)
Multiplying both sides by r gives:
mv² = Ze²/(4πε₀r)

• Bohr's Quantization Condition
Bohr proposed that the angular momentum of the electron is quantized:
mvr = nh/2π
Rearranging,
v = nh/(2πmr)

• Substitution
Substituting the value of v into the force equation and simplifying gives:
r = (ε₀h²n²)/(πme²Z)
Since
a₀ = ε₀h²/(πme²)
the equation becomes:
rₙ = (n²a₀)/Z
This is the required expression for Bohr's atomic radius.

➡️ Significance of Bohr's Atomic Radius
Bohr's atomic radius is significant because it provides a theoretical method for calculating the size of hydrogen and hydrogen-like atoms. It demonstrates that the radius of an electron's orbit increases with increasing principal quantum number, indicating that electrons farther from the nucleus occupy larger orbits. The concept also explains why excited atoms have larger sizes than atoms in their ground state. Furthermore, the Bohr radius forms the basis for understanding atomic structure, electron configuration, and the emission spectra of hydrogen-like atoms.

➡️Force Between the Nucleus and the Electron
The nucleus of an atom carries a positive charge due to the presence of protons, while the electron carries a negative charge. These opposite charges attract each other through an electrostatic force known as the Coulomb force. This attractive force is responsible for holding the electron in its orbit around the nucleus. In Bohr's atomic model, this electrostatic force acts as the centripetal force required for the circular motion of the electron.

• Formula for the Force
The electrostatic force between the nucleus and the electron is given by:
F = (1/4πε₀) × (Ze²/r²)
Where:
F = Electrostatic force
ε₀ = Permittivity of free space
Z = Atomic number
e = Electronic charge
r = Distance between the nucleus and the electron

This equation shows that the force is directly proportional to the atomic number and inversely proportional to the square of the distance between the nucleus and the electron.

• Derivation of the Force
The derivation begins with Coulomb's law, which states that the force between two charged particles is:
F = k(q₁q₂)/r²

where
k = 1/(4πε₀)
For a hydrogen-like atom:
Charge on the nucleus = +Ze
Charge on the electron = −e

Considering the magnitude of the charges,
F = k(Ze²)/r²
Substituting the value of k gives:
F = (1/4πε₀) × (Ze²/r²)

Since this electrostatic attraction keeps the electron moving in a circular orbit, it is equal to the centripetal force:
mv²/r = (1/4πε₀) × (Ze²/r²)
This relationship forms the basis for deriving Bohr's atomic radius and energy levels.

➡️ Significance of the Force Between the Nucleus and the Electron
The electrostatic force plays a fundamental role in atomic structure. It binds the electron to the nucleus and prevents it from escaping under normal conditions. The force also provides the necessary centripetal force that keeps the electron in a stable circular orbit. As the atomic number increases, the attractive force between the nucleus and the electron becomes stronger, resulting in smaller atomic radii for hydrogen-like atoms. This concept is essential for understanding atomic stability, ion formation, electron binding energy, and the arrangement of electrons within atoms.

ROCKET THRUST
Rocket thrust is the force that propels a rocket forward during flight. It is generated when fuel and an oxidizer react inside the rocket engine's combustion chamber to produce high-pressure, high-temperature gases. These gases are accelerated and expelled backward through a nozzle at very high speed. According to Newton's Third Law of Motion, every action has an equal and opposite reaction. Therefore, as the rocket expels gases backward, an equal and opposite force pushes the rocket forward. This forward force is known as rocket thrust. Unlike aircraft engines, rockets carry both fuel and oxidizer, allowing them to operate efficiently both within Earth's atmosphere and in the vacuum of space.

➡️ Rocket Thrust Equation
The thrust produced by a rocket is expressed by the equation:
F = ṁVe + (Pe − Pa)Ae

where:
F = Rocket thrust (N
ṁ = Mass flow rate of exhaust gases (kg/s)
Ve = Exhaust velocity at the nozzle exit (m/s)
Pe = Pressure of the exhaust gases at the nozzle exit (Pa)
Pa = Ambient atmospheric pressure (Pa)
Ae = Cross-sectional area of the nozzle exit (m²)

The first term, ṁVe, represents the momentum thrust, which results from the high-speed ejection of exhaust gases.

The second term, (Pe − Pa)Ae, represents the pressure thrust, which arises when the exhaust pressure differs from the surrounding atmospheric pressure. The total thrust is therefore the sum of these two components.

➡️ Derivation of the Rocket Thrust Equation
• Principle of Conservation of Momentum
The derivation of the rocket thrust equation is based on the principle of conservation of momentum and Newton's Second Law of Motion. Since force is defined as the rate of change of momentum, it can be written as:
F = d(mV)/dt

For a rocket ejecting gases continuously, the rate of mass flow is represented by
ṁ = dm/dt.

If the gases leave the nozzle with an exhaust velocity Ve, the force due to the momentum of the expelled gases becomes:
F = ṁVe
This component is called the momentum thrust because it results from the continuous change in momentum of the exhaust gases.

• Pressure Thrust
In addition to momentum thrust, another force is produced if the pressure of the exhaust gases at the nozzle exit is different from the surrounding atmospheric pressure. This pressure force is given by:
Fp = (Pe − Pa)Ae
where Pe is the exit pressure, Pa is the ambient pressure, and Ae is the nozzle exit area. This component is known as the pressure thrust.

• Total Rocket Thrust
The total thrust acting on the rocket is obtained by adding the momentum thrust and the pressure thrust. Therefore,
F = ṁVe + (Pe − Pa)Ae

If the nozzle is perfectly expanded so that the nozzle exit pressure equals the atmospheric pressure (Pe = Pa), the pressure thrust becomes zero because:
(Pe − Pa)Ae = 0

Under this condition, the thrust equation simplifies to:
F = ṁVe

Thus, only the momentum of the exhaust gases contributes to the rocket's thrust.

➡️ Significance of Rocket Thrust
1. Rocket thrust is one of the most important parameters in rocket propulsion because it determines the rocket's ability to move and perform its intended mission. First, it provides the force required to overcome gravity during launch. Without sufficient thrust, a rocket cannot lift off from the Earth's surface.

2. Second, rocket thrust determines the acceleration and velocity of the rocket throughout its flight. Higher thrust enables the rocket to reach greater speeds and altitudes.

3. Third, rocket thrust allows spacecraft to operate in outer space, where there is no atmosphere. Since rockets carry both fuel and oxidizer, they do not depend on atmospheric oxygen for combustion.

4. Fourth, thrust is essential for controlling spacecraft during orbital maneuvers, satellite deployment, interplanetary travel, and landing operations. Small adjustments in thrust enable precise navigation and positioning.

5. Finally, rocket thrust is a key factor in engine design and performance analysis. Engineers use the thrust equation to optimize nozzle geometry, exhaust velocity, and propellant flow rate in order to achieve maximum efficiency and payload capacity.

➡️ Problems and Solutions
Q1.
A rocket ejects exhaust gases at a mass flow rate of 80 kg/s with an exhaust velocity of 2,500 m/s. The nozzle exit pressure is equal to the atmospheric pressure. Calculate the rocket thrust.

Solution
Given:
ṁ = 80 kg/s
Ve = 2,500 m/s
Pe = Pa
Formula:
F = ṁVe + (Pe − Pa)Ae
Since Pe = Pa,
(Pe − Pa)Ae = 0
Therefore,
F = ṁVe

F = 80 × 2,500
F = 200,000 N
Answer: The rocket thrust is 200 kN.

Q2.
A rocket produces a thrust of 300,000 N while ejecting gases at 100 kg/s. Neglect pressure thrust.
Find the exhaust velocity.

Solution
Formula:
F = ṁVe

Therefore,
Ve = F/ṁ
Ve = 300,000/100
Ve = 3,000 m/s
Answer: Exhaust velocity = 3,000 m/s

Q3.
A rocket develops a thrust of 450,000 N with an exhaust velocity of 3,000 m/s. Neglect pressure thrust.
Calculate the mass flow rate.

Solution
Formula:
ṁ = F/Ve

ṁ = 450,000/3,000
ṁ = 150 kg/s
Answer: Mass flow rate = 150 kg/s

Q4.
A rocket develops a pressure thrust of 40,000 N through a nozzle with an exit area of 0.4 m². The atmospheric pressure is 100 kPa.
Determine the exit pressure.

Solution
Formula:
Pressure thrust = (Pe − Pa)Ae
Rearranging,
Pe = (Pressure thrust/Ae) + Pa

Pe = (40,000/0.4) + 100,000
Pe = 100,000 + 100,000
Pe = 200,000 Pa
Pe = 200 kPa
Answer: Exit pressure = 200 kPa