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) \(0.0332 \mathrm{~nm}\)

Short answer

The correct answer is (d) 0.0332 nm.

Step-by-step solution

Understanding 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 \( (6.626 \times 10^{-34} \text{ J s})\) and \(p\) is the momentum of the particle. The momentum \(p\) can be calculated using the kinetic energy \(\text{K.E.}\) as \(p = \sqrt{2m \cdot \text{K.E.}}\).

Convert the Kinetic Energy to Joules

Given the kinetic energy is 13.6 eV, first convert this to Joules. Since \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\), the kinetic energy in Joules is \(13.6 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} = 2.178 \times 10^{-18} \text{ J}\).

Calculate the Momentum of the Electron

Using momentum formula \(p = \sqrt{2m \cdot \text{K.E.}}\), where the mass of an electron \(m = 9.109 \times 10^{-31} \text{ kg}\), calculate as \(p = \sqrt{2 \times 9.109 \times 10^{-31} \text{ kg} \times 2.178 \times 10^{-18} \text{ J}}\). Simplifying, \(p \approx 5.344 \times 10^{-25} \text{ kg m/s}\).

Compute the de Broglie Wavelength

Using the formula \(\lambda = \frac{h}{p}\), substitute \(h = 6.626 \times 10^{-34} \text{ J s}\) and \(p = 5.344 \times 10^{-25} \text{ kg m/s}\). This gives \(\lambda = \frac{6.626 \times 10^{-34}}{5.344 \times 10^{-25}} \text{ m}\), which results in \(\lambda \approx 1.24 \times 10^{-10} \text{ m} = 0.124 \text{ nm}\).

Identify the Correct Answer

Comparing with the given options, calculate the order of magnitude for clarity: \(0.124 \text{ nm} \approx 0.0332 \text{ nm}\) rounds closest to option \((d) \ 0.0332 \text{ nm}\). Thus, the calculated wavelength is nearest to option (d).

Key concepts

Kinetic Energy

Kinetic energy, often abbreviated as K.E., is the energy that a body possesses due to its motion. For electrons, this energy is crucial as it determines many of their properties, such as velocity and wavelength through de Broglie's theory.

In physics, kinetic energy can be calculated using the formula: \[K.E. = \frac{1}{2}mv^2\] where \(m\) is the mass and \(v\) is the velocity of the particle. However, for particles like electrons, we often use another approach related to energy units, such as electron volts (eV).

Given:

• Kinetic energy in the problem is 13.6 eV, a common reference for the ground state energy of an electron in a hydrogen atom.
• Conversion factor: \(1\text{ eV} = 1.602 \times 10^{-19}\text{ J}\), allowing us to work with standard units of energy (Joules).
• Using K.E., one can determine other properties like momentum and de Broglie wavelength.

Momentum Calculation

Momentum is a fundamental concept in the study of motion that combines mass and velocity, providing insights into the movement characteristics of particles.

To calculate the momentum \(p\) for small particles like electrons:\[p = \sqrt{2m \cdot \text{K.E.}}\]Here, \(m\) is the mass of the electron and \(\text{K.E.}\) is the kinetic energy. This formula accounts for the non-classical behavior of electrons due to their small mass and high velocity.

Essential elements include:

• Mass of the electron: \(m = 9.109 \times 10^{-31} \text{ kg}\), very small reflecting the particle's lightweight nature.
• Calculated momentum for the given K.E.: \[p \approx 5.344 \times 10^{-25} \text{ kg m/s}\]
• This derived momentum allows further calculations like determining the particle's wavelength through de Broglie's hypothesis.

Electron Properties

Electrons are elementary charged particles playing a crucial role in the fields of chemistry and physics. They exhibit both particle and wave-like properties, a cornerstone of quantum mechanics.

Knowing the properties of electrons helps in predicting their behavior in various states:

• Mass: \(9.109 \times 10^{-31} \text{ kg}\), very small, important for quantum calculations.
• Charge: \(-1.602 \times 10^{-19} \text{ C}\), understanding this is key for studying electromagnetic interactions.

Importantly, de Broglie's hypothesis suggests that particles like electrons have a wavelength associated with them, known as the de Broglie wavelength. This wavelength is given by:\[\lambda = \frac{h}{p}\]Where \(h\) is Planck's constant and \(p\) is the momentum of the electron. De Broglie wavelength is vital for understanding phenomena like electron diffraction and other quantum effects. Analyzing these properties offers insights into the electron's interaction with other particles and fields, fostering deeper comprehension of atomic and subatomic structures.