Problem 118 - A Complete Resource Book in Chemistry for JEE Main

Question

Calculate the wavelength and energy of the radiation emitted for the electronic transition from infinity \((\infty)\) to stationary state first of the hydrogen atom. \(\left(\mathrm{R}_{\mathrm{H}}=1.09678 \times 10^{7} \mathrm{~m}^{-1}, \mathrm{~h}=6.6256 \times 10^{-34} \mathrm{Js}\right)\) (a) \(2.18 \times 10^{-21} \mathrm{~kJ}\) (b) \(3.18 \times 10^{-22} \mathrm{~kJ}\) (c) \(1.18 \times 10^{-23} \mathrm{~kJ}\) (d) \(2.18 \times 10^{-31} \mathrm{~kJ}\)

Step-by-Step Solution

Answer

Option (a) \(2.18 \times 10^{-21} \mathrm{~kJ}\).

Step 1: Understand the problem

We need to calculate both the wavelength and energy of radiation emitted during the electronic transition from infinity to the first stationary state of a hydrogen atom. The problem provides us with the Rydberg constant \( \mathrm{R}_{\mathrm{H}} = 1.09678 \times 10^{7} \mathrm{~m}^{-1} \) and Planck's constant \( \mathrm{h} = 6.6256 \times 10^{-34} \mathrm{Js} \).

Step 2: Apply the Rydberg Formula

The energy transition can be found using the Rydberg formula for the hydrogen atom, which is \( \Delta E = h \cdot c \cdot \mathrm{R}_{\mathrm{H}} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \), where \( n_1 = 1 \) and \( n_2 = \infty \).

Step 3: Simplify the Expression

Since \( n_2 = \infty \), the term \( \frac{1}{n_2^2} \) becomes 0. Thus, the equation simplifies to \( \Delta E = h \cdot c \cdot \mathrm{R}_{\mathrm{H}} \cdot \frac{1}{1^2} = h \cdot c \cdot \mathrm{R}_{\mathrm{H}} \).

Step 4: Substitute Known Values

Substitute the known values of \( h \), \( c \), and \( \mathrm{R}_{\mathrm{H}} \) into the simplified equation: \( \Delta E = 6.6256 \times 10^{-34} \cdot 3 \times 10^8 \cdot 1.09678 \times 10^7 \).

Step 5: Calculate Energy in Joules

Perform the multiplication to find the energy in joules: \( \Delta E = (6.6256 \times 10^{-34}) \cdot (3 \times 10^8) \cdot (1.09678 \times 10^7) = 2.1799 \times 10^{-18} \) J.

Step 6: Convert Joules to Kilojoules

To convert from joules to kilojoules, divide by 1000: \( 2.1799 \times 10^{-18} \text{ J} = 2.1799 \times 10^{-21} \text{ kJ} \).

Step 7: Identify the Correct Answer

Compare the calculated energy with the options provided. Hence, the energy of 2.1799 \times 10^{-21} kJ corresponds with option (a) \(2.18 \times 10^{-21} \mathrm{~kJ}\).

Key Concepts

Rydberg Formula

The Rydberg formula is a valuable tool in quantum mechanics used to predict the wavelengths of photons emitted during transitions between energy levels in a hydrogen atom. This formula is expressed as:

• \[ u = R_{\text{H}} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where \( u \) represents the frequency of the emitted or absorbed photon, \( R_{\text{H}} \) is the Rydberg constant, and \( n_1 \) and \( n_2 \) are the principal quantum numbers of the electron's initial and final states, respectively.
When an electron transitions to a lower energy level, it emits a photon with energy that corresponds to the difference in energy levels. This is a key concept, as energy is directly related to wavelength and frequency in electromagnetic radiation.

Energy of Radiation

Radiation energy plays a crucial role in understanding electronic transitions within atoms. The energy of an emitted photon results from electrons moving from a higher energy level to a lower one in the hydrogen atom, releasing this excess energy as a photon.
The energy difference \( \Delta E \) can be calculated using the Rydberg Formula by rearranging it:

• \[ \Delta E = h \cdot u = h \cdot c \cdot R_{\text{H}} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:

• \( h \) is Planck's constant.
• \( c \) is the speed of light.

By calculating \( \Delta E \), we can determine the amount of energy in joules per photon, which can then be converted to kilojoules by dividing by 1000.

Wavelength Calculation

The wavelength of radiation is a pivotal concept in describing electromagnetic waves. For a given transition, once you know the frequency \( u \), you can find the wavelength \( \lambda \) using the relationship between speed, frequency, and wavelength:

• \[ c = \lambda \cdot u \]

Where:

• \( c \) is the speed of light (\[ 3 \times 10^8\] m/s).

Solving for \( \lambda \), we get:

• \[ \lambda = \frac{c}{u} \]

This equation illustrates that wavelength and frequency are inversely proportional. A higher frequency corresponds to a shorter wavelength. Understanding this inverse relationship helps predict how different types of radiation will behave based on their source or interaction with matter.

Planck's Constant

Planck's constant is fundamental in the realm of quantum mechanics. It characterizes the size of the quantum of electromagnetic action that relates energy carried by a photon to its frequency. The constant is denoted by \( h \) and has a value of \[ 6.626 \times 10^{-34} \] Joule-seconds.
It appears in various equations such as the energy-frequency relation:

• \[ E = h \cdot u \]

Where:

• \( E \) is energy,
• \( h \) is Planck's constant,
• \( u \) is frequency.

Planck's constant essentially links the macro world of classical physics with the quantum world. It highlights that energy changes occur in discrete amounts, known as quanta, offering a deeper insight into atomic and subatomic processes.