Problem 10 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Heat treatment of muscular pain involves radiation of wavelength of about 900 \(\mathrm{nm}\). Which spectral line of \(\mathrm{H}\)-atom is suitable for this purpose? [Main Jan. 11, 2019 (I)] \(\left[\mathrm{R}_{\mathrm{H}}=1 \times 10^{5} \mathrm{~cm}^{-1} . \mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\right]\) (a) Paschen, \(\infty \rightarrow 3\) (b) Paschen, \(5 \rightarrow 3\) (c) Balmer, \(\infty \rightarrow 2\) (d) Lyman, \(\infty \rightarrow 1\)

Step-by-Step Solution

Answer

Option (b) Paschen, \(5 \rightarrow 3\) is suitable.

Step 1: Convert Wavelength to Frequency

Given the wavelength \( \lambda = 900 \, \text{nm} \), convert it to meters: \( 900 \, \text{nm} = 900 \times 10^{-9} \, \text{m} \). The frequency \( u \) is calculated using the formula \( u = \frac{c}{\lambda} \), where \( c = 3 \times 10^8 \, \text{m/s} \). Substituting the values, we get \( u = \frac{3 \times 10^8}{900 \times 10^{-9}} \approx 3.33 \times 10^{14} \, \text{Hz} \).

Step 2: Calculate the Corresponding Energy

Energy \( E \) of a photon is given by \( E = h u \), where \( h = 6.6 \times 10^{-34} \, \text{Js} \) is Planck's constant. Substituting the frequency from Step 1, \( E = 6.6 \times 10^{-34} \times 3.33 \times 10^{14} \approx 2.2 \times 10^{-19} \, \text{J} \).

Step 3: Convert Energy to Wavenumber

First, convert the energy into electron volts (eV). Using \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \), we find \( E \approx 2.2 \times 10^{-19} \, \text{J} = \frac{2.2 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.375 \, \text{eV} \). In terms of wavenumber, \( \text{wavenumber} = \frac{E}{hc} \approx \frac{1.375}{1240} \approx 0.0011 \, \text{nm}^{-1} \).

Step 4: Identify the Suitable Spectral Line

Using the Rydberg formula for hydrogen \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \), convert \( 0.0011 \, \text{nm}^{-1} = 11000 \, \text{cm}^{-1} \). Solving for \( n_1 \) and \( n_2 \), and using the series: Lyman (to \( n_1 = 1 \)), Balmer (to \( n_1 = 2 \)), Paschen (to \( n_1 = 3 \)), we find \( n_1 = 3 \) matches, so choose the Paschen series, particularly option (b) \( 5 \rightarrow 3 \).

Key Concepts

Electromagnetic Spectrum

The electromagnetic spectrum encompasses all types of electromagnetic radiation, each characterized by a unique wavelength or frequency. From the longest wavelength with the lowest energy to the shortest wavelength with the highest energy, the spectrum includes radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays.

Infrared radiation, part of the spectrum just beyond visible light, plays a pivotal role in the heat treatment of muscular pain because of its ability to penetrate the skin and provide warmth. Wavelengths like 900 nm, which fall within the infrared domain, are specifically utilized in therapeutic treatments.

• Infrared: 700 nm to 1 mm, often used in medical therapies.
• Visible light: 380-750 nm, primarily for human vision.
• Ultraviolet: 10-400 nm, can cause skin damage.

Understanding the electromagnetic spectrum helps in selecting the right type of radiation, such as the infrared region in our exercise, ideal for pain alleviation.

Photon Energy Calculation

Calculating the energy of a photon involves using the relationship between its frequency and Planck's constant. The energy equation is given by \( E = h u \), where \( E \) is the energy, \( h \) is Planck's constant (\( 6.6 \times 10^{-34} \, \text{Js} \)), and \( u \) is the frequency of the light.

When dealing with the wavelength, as given in the problem, it's crucial to first convert it to frequency using the speed of light \( c = 3 \times 10^8 \, \text{m/s} \) through the formula \( u = \frac{c}{\lambda} \). For example, a wavelength of \( 900 \, \text{nm} \) translates to a frequency approximately \( 3.33 \times 10^{14} \, \text{Hz} \), leading to photon energy around \( 2.2 \times 10^{-19} \, \text{J} \).

• Wavelength \( \lambda = 900 \, \text{nm} \) \( \Rightarrow \) Frequency \( u \)
• Frequency \( u = 3.33 \times 10^{14} \, \text{Hz} \) \( \Rightarrow \) Energy \( E \)

This methodology allows for understanding how energy levels relate to specific wavelengths in various applications.

Rydberg Formula

The Rydberg Formula is instrumental in calculating the spectral lines of hydrogen and predicting the wavelengths emitted or absorbed during electron transitions between energy levels. Using the formula \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \), we can find wavelengths associated with specific transitions:

• \( R_H = 1.097 \times 10^7 \, \text{m}^{-1} \) is the Rydberg constant for hydrogen.
• \( n_1 \) and \( n_2 \) represent the lower and higher energy levels, respectively.
• The series are named based on \( n_1 \): Lyman (\( n_1 = 1 \)), Balmer (\( n_1 = 2 \)), Paschen (\( n_1 = 3 \)), etc.

In the context of our exercise, the transition matching a wavenumber of \( 11000 \, \text{cm}^{-1} \) suggests the usage of the Paschen series specifically \( n_2 = 5 \rightarrow n_1 = 3 \). Understanding these transitions is crucial for recognizing the appropriate spectral line, such as when selecting suitable wavelengths for different applications like medical treatments.