Problem 14 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The radius of the second Bohr orbit for hydrogen atom is : (Plank's const. \(h=6.6262 \times 10^{-34} \mathrm{Js}\); mass of electron \(=9.1091 \times 10^{-31} \mathrm{~kg}\); charge of electron \(\mathrm{e}=1.60210 \times 10^{-19} \mathrm{C}\); permittivity of vaccum \(\left.\epsilon_{0}=8.854185 \times 10^{-12} \mathrm{~kg}^{-1} \mathrm{~m}^{-3} \mathrm{~A}^{2}\right)\) (a) \(1.65 \AA\) (b) \(4.76 \mathrm{~A}\) (c) \(0.529 \AA\) (d) \(2.12 \AA\)

Step-by-Step Solution

Answer

The radius of the second Bohr orbit for hydrogen is 2.12 Ångströms.

Step 1: Understand the Bohr's radius formula

The formula for the radius of the nth orbit of hydrogen is given by \[ r_n = \frac{n^2 h^2}{4\pi^2 m e^2 \varepsilon_0} \]where \( n \) is the orbit number, \( h \) is Planck's constant, \( m \) is the electron's mass, \( e \) is the electron's charge, and \( \varepsilon_0 \) is the permittivity of vacuum. We must solve for the second orbit, so \( n = 2 \).

Step 2: Substitute the given values into the formula

Substitute \( n = 2 \), \( h = 6.6262 \times 10^{-34} \) \( J \cdot s \), \( m = 9.1091 \times 10^{-31} \) \( kg \), \( e = 1.60210 \times 10^{-19} \) \( C \), and \( \varepsilon_0 = 8.854185 \times 10^{-12} \) \( kg^{-1} \cdot m^{-3} \cdot A^2 \) into the formula:\[ r_2 = \frac{2^2 (6.6262 \times 10^{-34})^2}{4\pi^2 (9.1091 \times 10^{-31}) (1.60210 \times 10^{-19})^2 (8.854185 \times 10^{-12})} \]

Step 3: Calculate the radius of the second orbit

Now, compute the value of the expression:- Calculate the numerator: \( 4 \times (6.6262 \times 10^{-34})^2 \)- Calculate the denominator: \( 4\pi^2 \times (9.1091 \times 10^{-31}) \times (1.60210 \times 10^{-19})^2 \times (8.854185 \times 10^{-12}) \)- Divide the two results to get \( r_2 \).

Step 4: Convert result to Angstroms

After calculating \( r_2 \), remember that 1 meter = 10^{10} Ångströms. Convert your result from meters to Ångströms, and compare with the given options.

Step 5: Determine the correct answer

After performing the calculation and conversion, you will find that the radius of the second Bohr orbit for the hydrogen atom is approximately 2.12 Ångströms.

Key Concepts

Hydrogen Atom

The hydrogen atom is the simplest atom and is composed of just one proton and one electron. In models of the atom, such as that proposed by Niels Bohr, it serves as the baseline for understanding atomic structure in greater detail.
Bohr's model of the atom helps to explain how electrons are arranged in discrete energy levels or "orbits" around the nucleus. This is a pivotal concept in physics and has been integral in the development of modern quantum mechanics.
In the hydrogen atom, the electron orbits the single proton in circular paths. The energy levels of these orbits are quantized, meaning the electron can only occupy specific energy levels. This quantization results in the emission or absorption of light or electromagnetic radiation, leading to the formation of line spectra, a characteristic feature of different elements.

Bohr Radius

The Bohr radius is a fundamental physical constant and represents the most probable distance between the nucleus and the electron in a hydrogen atom in its ground state.
Mathematically, the Bohr radius can be calculated using the formula:

• \[ r_n = \frac{n^2 h^2}{4\pi^2 m e^2 \varepsilon_0} \]

• \(n\) denotes the principal quantum number signifying the electron's orbit.
• \(h\) is Planck's constant.
• \(m\) stands for the electron mass.
• \(e\) is the charge of an electron.
• \(\varepsilon_0\) represents the permittivity of vacuum.

The Bohr radius is essential in calculations involving atomic structures, like determining the size of atomic orbits, effective field strengths, and potential energies.
Its standard value is approximately 0.529 Ångströms, characterizing the scale of atomic and molecular distances.

Quantum Mechanics

Quantum Mechanics is the fundamental theory in physics that describes nature at the smallest scales, such as atomic and subatomic levels.
This field introduces the idea that particles such as electrons also exhibit wave-like properties, leading to the concept of wave-particle duality. The understanding of electron behavior in atoms was revolutionized by applying quantum mechanics because it allowed for the prediction of discrete energy levels within atoms.
Unlike classical physics, which views electron paths as predictable orbits, quantum mechanics describes them as probability distributions. The position, momentum, and energy of an electron are understood in terms of probabilities rather than definite paths.
Quantum mechanics also extends beyond the hydrogen atom to inform our understanding of chemical bonding, electronic structures of atoms, and more complex molecules, substantially influencing modern technology and materials science.