Problem 57 - A Complete Resource Book in Chemistry for JEE Main

Question

The de Broglie wavelength of the electron in the ground state of hydrogen atom is \([\mathrm{K} . \mathrm{E}=13.6 \mathrm{eV}]\); \(\mathrm{leV}=1.602 \times 10^{-19} \mathrm{~J}\) (a) \(33.28 \mathrm{~nm}\) (b) \(3.328 \mathrm{~nm}\) (c) \(0.3328 \mathrm{~nm}\) (d) \(0.0332 \mathrm{~nm}\)

Step-by-Step Solution

Answer

The correct answer is (c) \(0.3328\,\text{nm}\).

Step 1: Understand the de Broglie wavelength formula

The de Broglie wavelength \( \lambda \) is given by the formula: \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ m}^2 \text{ kg s}^{-1} \), and \( p \) is the momentum of the particle. For an electron, its momentum \( p = mv \), where \( m \) is the mass and \( v \) is the velocity.

Step 2: Relate momentum to kinetic energy

The kinetic energy (K.E.) of an electron is given as \( 13.6 \text{ eV} \). We convert this to joules: \( \text{K.E.} = 13.6 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} = 2.179 \times 10^{-18} \text{ J} \). The kinetic energy is also \( \frac{1}{2}mv^2 \) which gives the velocity \( v = \sqrt{\frac{2 \cdot \text{K.E.}}{m}} \).

Step 3: Obtain momentum from kinetic energy

Express momentum \( p \) in terms of \( v \) and \( m \) as \( p = mv \). Using \( v = \sqrt{\frac{2 \cdot K.E.}{m}} \), substitute into \( p = mv \) to get \( p = m \sqrt{\frac{2\cdot \text{K.E.}}{m}} = \sqrt{2m \cdot \text{K.E.}} \).

Step 4: Substitute values into the de Broglie equation

Substitute \( p = \sqrt{2m \cdot \text{K.E.}} \) into the de Broglie wavelength formula \( \lambda = \frac{h}{p} \). Thus, \( \lambda = \frac{h}{\sqrt{2m \cdot \text{K.E.}}} \), where \( m = 9.109 \times 10^{-31} \text{ kg} \), \( h = 6.626 \times 10^{-34} \text{ m}^2 \text{ kg/s} \), and \( \text{K.E.} = 2.179 \times 10^{-18} \text{ J} \).

Step 5: Calculate the de Broglie wavelength

Insert the values into the formula: \( \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \cdot 9.109 \times 10^{-31} \cdot 2.179 \times 10^{-18}}} \approx 0.3328 \text{ nm} \).

Step 6: Choose the correct option

The calculated de Broglie wavelength of the electron is approximately \( 0.3328 \text{ nm} \), hence the correct answer is option (c).

Key Concepts

Hydrogen Atom

The hydrogen atom is the simplest atom, consisting of just one proton and one electron. Serving as a fundamental model in physics and chemistry, it helps us understand atomic behavior and structure. In the context of de Broglie wavelength calculations, the hydrogen atom's electron in its ground state implements the basic principles of quantum mechanics.

• It provides a model for electron behavior around an atomic nucleus.
• The ground state is when the electron is closest to the nucleus, having the least energy.
• This low-energy state is the basis for calculations of properties like wavelength and momentum.

Recognizing that the hydrogen atom can be analyzed using quantum physics principles reflects the wider application of these concepts in understanding more complex atoms.

Kinetic Energy Conversion

Kinetic energy conversion is vital in understanding how energy states change within an electron. The kinetic energy (K.E.) of an electron is related to its motion, quantified by its velocity. In the exercise, we focused on converting the kinetic energy from electronvolts (eV) to joules (J) to carry out necessary calculations.

• Converting units: 1 eV is equal to \(1.602 \times 10^{-19}\) J. Hence, converting 13.6 eV provides clarity when using SI units in formulas.
• The relationship of K.E. to velocity: K.E. is calculated by \(\frac{1}{2}mv^2\), where \(m\) is the mass of the electron. Our goal is to extract velocity for further calculations.
• Knowing velocity: It's derived as \(v = \sqrt{\frac{2 \cdot \text{K.E.}}{m}}\), where \(m\) is known, when K.E. is converted into joules.

This fundamental understanding allows transitions between energy and motion properties leading to other calculations.

Planck's Constant

Planck's constant \(h\) is a critical physical constant that applies across quantum mechanics. It reflects the proportionality between the energy of a photon and the frequency of its electromagnetic wave. In our exercise, this constant is essential for calculating the de Broglie wavelength.

• Planck's constant value is \(6.626 \times 10^{-34} \text{ m}^2 \text{ kg s}^{-1}\).
• It provides the link between de Broglie wavelength \(\lambda = \frac{h}{p}\) and momentum \(p\) of particles.
• Its unique property allows us to traverse between microscopic quantum scales and macroscopic classical scales.

Using Planck's constant ensures that our calculations are consistent and accurate, linking energy, frequency, and wavelength in a coherent framework.

Electron Momentum

Electron momentum, denoted as \(p\), illustrates the quantity of motion an electron possesses and is key in determining the de Broglie wavelength. Momentum links the particle-like nature of electrons to wave-like properties in quantum mechanics.

• Expressed as \(p = mv\), where \(m\) is the mass, and \(v\) is the velocity derived from kinetic energy.
• Directly used with the de Broglie formula, \(\lambda = \frac{h}{p}\), to find the wavelength associated with particle motion.
• Facilitates understanding of how an electron exhibits both wave and particle characteristics, supporting the duality principle in quantum physics.

Momentum's role in quantum physics reinforces the connection between classical mechanics and quantum observations, showing the continuous nature of these fields.