Problem 44 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The energy of the electron in the second and the third Bohr's orbits of the hydrogen atom is \(-5.42 \times 10^{-12}\) erg and \(-2.41 \times 10^{-12}\) erg respectively. Calculate the wavelength of the emitted radiation when the electron drops from the third to the second orbit.

Step-by-Step Solution

Answer

The wavelength of the emitted radiation is approximately 6590 nm.

Step 1: Identify the Energy Levels

The problem states the energies as \(-5.42 \times 10^{-12}\) erg for the second Bohr orbit and \(-2.41 \times 10^{-12}\) erg for the third orbit.

Step 2: Calculate the Energy Change

The energy change, \(\Delta E\), when the electron transitions from the third orbit to the second is calculated by subtracting the energy of the second orbit from the third: \[ \Delta E = E_3 - E_2 = (-2.41 \times 10^{-12}) - (-5.42 \times 10^{-12}) \] This simplifies to:\[ \Delta E = 3.01 \times 10^{-12} \text{ erg} \]

Step 3: Use Energy-Wavelength Relation

The relationship between energy change and wavelength \(\lambda\) of emitted radiation is given by Planck's equation:\[ \Delta E = \frac{hc}{\lambda} \]where \(h\) is Planck's constant \(6.626 \times 10^{-27}\) erg·s and \(c\) is the speed of light \(3 \times 10^{10}\) cm/s.

Step 4: Solve for Wavelength

Rearranging the formula to solve for \(\lambda\):\[ \lambda = \frac{hc}{\Delta E} \]Substitute the values:\[ \lambda = \frac{(6.626 \times 10^{-27}) \times (3 \times 10^{10})}{3.01 \times 10^{-12}} \] Calculate to find:\[ \lambda \approx 6.59 \times 10^{-5} \text{ cm} \]

Step 5: Convert to More Common Units

Converting the wavelength from cm to nm (1 cm = 10 million nm), we get:\[ \lambda \approx 6590 \text{ nm} \]

Key Concepts

Energy Levels in Bohr's Model

In Bohr's model of the hydrogen atom, electrons travel in fixed circular paths or orbits with quantized energy levels. Each energy level is associated with a specific orbit and is characterized by the principal quantum number "n".
For hydrogen, the energy of an electron is given by: \[ E_n = - \frac{R_h}{n^2} \] where \( R_h \) is the Rydberg constant \((2.18 \times 10^{-18} \text{J})\), and \( n \) is the principal quantum number.
- **Second Orbit:** In our example, the second energy level \((n = 2)\) has an energy of \(-5.42 \times 10^{-12}\) erg.- **Third Orbit:** The third energy level \((n = 3)\) has an energy of \(-2.41 \times 10^{-12}\) erg.When an electron transitions between these two levels, it releases or absorbs a precise amount of energy, which is key to understanding the emission spectrum of hydrogen.

Wavelength Calculation and Energy Change

The change in energy \(\Delta E\) when an electron moves from one energy level to another is crucial for determining the wavelength of the emitted radiation. The process involves calculating \(\Delta E\), applying Planck's equation, and finally solving for wavelength.
To find \(\Delta E\), subtract the energy of the final state \((E_2)\) from the initial state \((E_3)\):\[ \Delta E = E_3 - E_2 = (-2.41 \times 10^{-12}) - (-5.42 \times 10^{-12})\] This results in \[ \Delta E = 3.01 \times 10^{-12} \text{ erg}\] indicating absorbed or released energy. By using Planck’s relation \(\Delta E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant and \(c\) is the speed of light, you can rearrange the formula to solve for the wavelength \(\lambda\):\[ \lambda = \frac{hc}{\Delta E} \] Substitute known values and solve to find the wavelength as \[ \lambda \approx 6.59 \times 10^{-5} \text{ cm}\] By converting this to nanometers:1 cm = 10 million nm, we find \(\lambda \approx 6590 \text{ nm}\).

Understanding Planck's Equation

Planck's equation, \[ \Delta E = \frac{hc}{\lambda} \] plays a pivotal role in connecting the concepts of energy and light in quantum mechanics. It establishes the direct relationship between the change in energy \((\Delta E)\) and the wavelength \((\lambda)\) of the radiation. This connection is vital in estimating wavelengths when electrons transition between energy levels.
- **Planck's Constant (\(h\))**: This is a fundamental constant in physics, \(6.626 \times 10^{-27}\) erg·s, signifying the smallest action in quantum mechanics.- **Speed of Light (\(c\))**: Always considered to be \(3 \times 10^{10}\) cm/s in formulas involving radiation.By using Planck's equation, one can derive the wavelength given a transition energy, emphasizing how energy quantization in Bohr's model elucidates the wave-particle duality of light. Understanding this equation helps in unraveling the principles of emission and absorption spectroscopy, fundamental in atomic physics.