Development Leading to the Bohr's Model of Atom

Introduction to Bohr's Model Foundation

Historically, the study of how radiation interacts with matter provided immense information regarding the structure of atoms and molecules. Neils Bohr utilized these experimental results to improve upon the atomic model proposed by Ernest Rutherford.

Two major developments played a pivotal role in the formulation of Bohr's model of the atom:

1. Dual character of electromagnetic radiation: The realization that radiations possess both wave-like and particle-like properties.
2. Experimental results regarding atomic spectra: Observations of the specific wavelengths of light emitted or absorbed by atoms.

Before diving into Bohr's model, we must first understand the dual nature of electromagnetic radiation, starting with its wave nature.

Wave Nature of Electromagnetic Radiation

In the mid-nineteenth century, physicists actively studied the absorption and emission of radiation by heated objects, known as thermal radiations. Today, it is a well-known fact that thermal radiations consist of electromagnetic waves of various frequencies or wavelengths.

Maxwell's Electromagnetic Theory

The first comprehensive explanation of the interaction between charged bodies and the behavior of electrical and magnetic fields on a macroscopic level was given by James Clerk Maxwell in 1870.

Maxwell suggested that when an electrically charged particle moves under acceleration, alternating electrical and magnetic fields are produced and transmitted. These fields are transmitted in the form of waves called electromagnetic waves or electromagnetic radiation.

Historical Context: Light is a form of radiation known since ancient times. Isaac Newton originally supposed light was made of particles (corpuscules). It was only in the 19th century that the wave nature of light was firmly established.

What is an Electromagnetic Wave?

An electromagnetic wave consists of oscillating electric and magnetic fields that are:

• Perpendicular to each other
• Perpendicular to the direction of wave propagation
• In phase (reach their maxima and minima at the same time)

Unlike sound waves, electromagnetic waves do not need a medium — they can travel through vacuum. The speed of all electromagnetic radiation in vacuum is the same:

$$c = 3 \times 10^8 \text{ m s}^{-1}$$

This is the speed of light, one of the fundamental constants of nature.

Properties of Electromagnetic Waves

Although electromagnetic wave motion is complex, we focus on a few simple, fundamental properties:

1. Perpendicular Fields: The oscillating electric and magnetic fields produced by oscillating charged particles are perpendicular to each other. Furthermore, both are perpendicular to the direction of propagation of the wave.
2. No Medium Required: Unlike sound waves or water waves, electromagnetic waves do not require a physical medium to propagate. They can move freely in a vacuum.
3. Electromagnetic Spectrum: There are many types of electromagnetic radiations, differing from one another in wavelength (or frequency). Together, they constitute the electromagnetic spectrum.

[School Exam Focus] You may be asked to list the properties of electromagnetic waves. Remember the three key points: perpendicular fields, vacuum propagation, and varying wavelengths/frequencies.

Characteristics of a Wave

A wave is characterised by the following properties:

1. Wavelength ($\lambda$)

The distance between two consecutive crests (or troughs) of a wave. It is measured in metres (m), but commonly expressed in:

• nanometre (nm): $1 \text{ nm} = 10^{-9}$ m
• picometre (pm): $1 \text{ pm} = 10^{-12}$ m
• ångström (Å): $1 \text{ Å} = 10^{-10}$ m

2. Frequency ($\nu$)

The number of waves passing a given point per second. It is measured in hertz (Hz) or s$^{-1}$.

• $1 \text{ Hz} = 1 \text{ s}^{-1}$ = one cycle per second

3. Amplitude ($A$)

The height of the crest (or depth of the trough) from the centre line. It determines the intensity or brightness of radiation.

4. Wave Number ($\bar{\nu}$)

The number of wavelengths per unit length: $$\bar{\nu} = \frac{1}{\lambda}$$ Unit: m$^{-1}$ or cm$^{-1}$ (commonly used in spectroscopy).

5. Velocity ($c$)

The distance travelled by a wave in one second. For electromagnetic radiation in vacuum: $$c = 3 \times 10^8 \text{ m s}^{-1}$$

The Fundamental Wave Equation

The relationship between speed, wavelength, and frequency is:

$$\boxed{c = \nu \lambda}$$

where:

• $c$ = speed of light ($3 \times 10^8$ m/s)
• $\nu$ = frequency (Hz)
• $\lambda$ = wavelength (m)

This equation tells us that frequency and wavelength are inversely proportional — as one increases, the other decreases.

We can also relate wave number to frequency: $$\bar{\nu} = \frac{1}{\lambda} = \frac{\nu}{c}$$

[JEE Tip] Always ensure consistent units before calculation. If $\lambda$ is given in nm, convert to metres: multiply by $10^{-9}$. If $\bar{\nu}$ is in cm$^{-1}$, convert to m$^{-1}$: multiply by $100$.

Key Point: All electromagnetic radiation travels at the same speed in vacuum, but different types of radiation differ in their wavelength and frequency.

The Electromagnetic Spectrum and Wave Parameters

The electromagnetic spectrum encompasses all types of electromagnetic radiation. Different regions of the spectrum are identified by different names based on their frequency ($\nu$) or wavelength ($\lambda$).

Key Regions of the Electromagnetic Spectrum

Region	Approximate Frequency (Hz)	Common Application
Radio Frequency	$\sim 10^6$	Broadcasting (AM/FM)
Microwave	$\sim 10^{10}$	Radar and heating
Infrared (IR)	$\sim 10^{13}$	Heating
Visible Light	$\sim 10^{15}$	The only part our eyes can detect
Ultraviolet (UV)	$\sim 10^{16}$	Component of sun's radiation

Note on Visible Light: The visible spectrum is a very small portion of the overall spectrum, ranging from 400 nm (violet) to 750 nm (red).

Particle Nature of Electromagnetic Radiation

While the wave nature of electromagnetic radiation successfully explains phenomena like diffraction (bending of waves around an obstacle) and interference (combination of waves), classical 19th-century physics failed to explain several crucial observations:

1. The nature of emission of radiation from hot bodies (black-body radiation).
2. Ejection of electrons from a metal surface when radiation strikes it (photoelectric effect).
3. Variation of heat capacity of solids as a function of temperature.
4. Line spectra of atoms, with special reference to hydrogen.

These phenomena indicate that a system can take energy only in discrete amounts, rather than continuously.

Black-Body Radiation

A black body is an ideal body that absorbs and emits all frequencies of radiation. When heated, a black body emits radiation whose distribution depends only on its temperature, not on its material.

• As temperature increases, a higher proportion of short wavelength (e.g., blue light) is generated.
• Example: An iron rod heated in a furnace turns dull red, then progressively more red, then white, and finally blue at very high temperatures.

Experimental observations:

• At a given temperature, the intensity of radiation increases with frequency, reaches a maximum, and then decreases at higher frequencies
• As temperature increases, the peak shifts to higher frequencies (shorter wavelengths)
• The total energy emitted increases with temperature

The problem: Classical wave theory predicted that the intensity should keep increasing indefinitely at higher frequencies — the so-called ultraviolet catastrophe. This prediction clearly contradicted experimental results.

Key Point: The failure to explain black body radiation using classical wave theory forced physicists to rethink the nature of radiation itself.

An ideal body that emits and absorbs radiations of all frequencies uniformly is called a black body, and the radiation it emits is black body radiation.

• In practice, a perfect black body doesn't exist, but carbon black is a close approximation.
• A good physical model is a cavity with a tiny hole; any radiation entering the hole is reflected internally until completely absorbed.

Intensity vs. Wavelength Graph: At a given temperature, the intensity of radiation emitted increases with wavelength, reaches a maximum value, and then decreases. As temperature increases, the maximum of the curve shifts to a shorter wavelength. Classical wave theory could not explain this curve.

Planck's Quantum Theory (1900)

Max Planck resolved the black body problem by proposing a revolutionary idea:

Energy is not emitted or absorbed continuously but in discrete packets called quanta (singular: quantum).

Key Postulates

1. Radiation is emitted or absorbed in the form of small packets of energy called quanta (or photons for light)
2. The energy of each quantum is proportional to its frequency:

$$\boxed{E = h\nu}$$

where:

• $E$ = energy of one quantum (J)
• $h$ = Planck's constant = $6.626 \times 10^{-34}$ J s
• $\nu$ = frequency (Hz)

1. Energy can only be emitted or absorbed in whole-number multiples of $h\nu$: $$E = nh\nu \quad (n = 1, 2, 3, \ldots)$$

This means energy is quantised — it comes in definite, discrete amounts, not in any arbitrary value.

Expressing Energy in Terms of Wavelength

Since $\nu = c/\lambda$: $$E = h\nu = \frac{hc}{\lambda}$$

This shows that shorter wavelength = higher energy.

[JEE Tip] Memorise the combined constant: $hc = 6.626 \times 10^{-34} \times 3 \times 10^8 = 1.988 \times 10^{-25}$ J m. In eV: $hc = 1240$ eV·nm (very useful for quick calculations!).

The Photoelectric Effect

What is the Photoelectric Effect?

When light of suitable frequency strikes a metal surface, electrons are ejected from the surface. These ejected electrons are called photoelectrons, and the phenomenon is called the photoelectric effect.

This was first observed by Heinrich Hertz (1887) and studied in detail by Hallwachs and Lenard.

Key Experimental Observations

1. Threshold frequency ($\nu_0$): For each metal, there is a minimum frequency below which no electrons are ejected, regardless of the intensity of light. This minimum frequency is called the threshold frequency.

2. Effect of frequency: If $\nu > \nu_0$, the kinetic energy of ejected electrons increases linearly with frequency. It does NOT depend on intensity.

3. Effect of intensity: Increasing the intensity of light (at $\nu > \nu_0$) increases the number of electrons ejected (i.e., the photocurrent) but does NOT change their kinetic energy.

4. Instantaneous emission: Electrons are ejected almost instantaneously (within $10^{-9}$ s), with no time lag.

Why Wave Theory Fails

According to wave theory:

• Energy depends on intensity (amplitude²), not frequency
• Electrons should be ejected at any frequency if intensity is high enough
• There should be a time delay for low-intensity light

All three predictions are wrong! Wave theory cannot explain the photoelectric effect.

Einstein's Explanation (1905)

Albert Einstein explained the photoelectric effect using Planck's quantum theory:

Light consists of particles called photons. Each photon carries energy $E = h\nu$. When a photon strikes a metal surface, its entire energy is transferred to a single electron.

Einstein's Photoelectric Equation

$$\boxed{h\nu = h\nu_0 + \frac{1}{2}m_e v^2}$$

or equivalently:

$$h\nu = W_0 + KE_{\max}$$

where:

• $h\nu$ = energy of the incident photon
• $h\nu_0 = W_0$ = work function (minimum energy needed to eject an electron)
• $\frac{1}{2}m_e v^2 = KE_{\max}$ = maximum kinetic energy of the ejected electron

Understanding the Equation

• If $\nu < \nu_0$: photon energy < work function → no electron ejected
• If $\nu = \nu_0$: photon energy = work function → electron ejected with zero KE
• If $\nu > \nu_0$: excess energy appears as kinetic energy of the electron

The Work Function

The work function ($W_0 = h\nu_0$) is a property of the metal:

Metal	Work Function (eV)	Threshold Wavelength (nm)
Caesium (Cs)	2.14	579
Potassium (K)	2.30	539
Sodium (Na)	2.75	451
Copper (Cu)	4.65	267
Platinum (Pt)	5.65	220

[NEET Tip] Einstein received the Nobel Prize in Physics (1921) for explaining the photoelectric effect, NOT for relativity!

Dual Nature of Radiation

Light shows wave nature (interference, diffraction) and particle nature (photoelectric effect, black body radiation). This is called the dual nature of electromagnetic radiation.

Light is neither purely a wave nor purely a particle — it behaves as both, depending on the experiment.

Memory Capsule

Key Concepts Summary:

• Maxwell's Theory: Accelerating charges produce perpendicular oscillating electric and magnetic fields (EM waves).
• Wave Parameters: $c = \nu \lambda$ and $\bar{\nu} = \frac{1}{\lambda}$.
• Visible Spectrum: 400 nm (violet) to 750 nm (red).
• Wave Theory Successes: Explains diffraction and interference.
• Wave Theory Failures: Fails to explain black-body radiation, photoelectric effect, heat capacity of solids, and line spectra.
• Planck's Quantum Theory: Energy is quantized. $E = h\nu$.
• Black Body: Perfect absorber and emitter. Peak intensity shifts to shorter $\lambda$ at higher temperatures.
• Photoelectric Effect: Instantaneous ejection of electrons from metals like K, Rb, Cs when light strikes.

Quick-Revision Formula Sheet:

Parameter	Formula	SI Unit	Common Unit
Speed of Light	$c = \nu \lambda$	$\text{m s}^{-1}$	$\text{m s}^{-1}$
Wavenumber	$\bar{\nu} = \frac{1}{\lambda}$	$\text{m}^{-1}$	$\text{cm}^{-1}$
Energy of Quantum	$E = h\nu$	$\text{J}$	$\text{J}$
Energy via Wavelength	$E = \frac{hc}{\lambda}$	$\text{J}$	$\text{J}$

Crucial Constants:

• Speed of light in vacuum ($c$): $3.0 \times 10^8 \text{ m s}^{-1}$
• Planck's constant ($h$): $6.626 \times 10^{-34} \text{ J s}$

Example 1: Calculating Wavelength from Frequency (Book Problem)

The Vividh Bharati station of All India Radio, Delhi, broadcasts on a frequency of 1,368 kHz (kilo hertz). Calculate the wavelength of the electromagnetic radiation emitted by transmitter. Which part of the electromagnetic spectrum does it belong to?

Solution:

1. Identify Given Information:

• Frequency ($\nu$) = $1368 \text{ kHz}$
• Speed of light ($c$) = $3.00 \times 10^8 \text{ m s}^{-1}$

1. Convert Units to SI:

• $\nu = 1368 \times 10^3 \text{ Hz} = 1368 \times 10^3 \text{ s}^{-1}$

1. State the Formula: The wavelength, $\lambda$, is related to frequency and speed of light by the equation: $$c = \nu \lambda \implies \lambda = \frac{c}{\nu}$$

2. Perform the Calculation: $$\lambda = \frac{3.00 \times 10^8 \text{ m s}^{-1}}{1368 \times 10^3 \text{ s}^{-1}}$$ $$\lambda = 219.3 \text{ m}$$

3. Identify the Spectrum Region: A wavelength of $219.3 \text{ m}$ is very long, which is a characteristic radiowave wavelength.

Takeaway: Always convert prefixes like 'kilo' ($10^3$) into standard scientific notation before dividing by the speed of light to avoid order-of-magnitude errors.

Example 2: Converting Wavelength Range to Frequency Range (Book Problem)

The wavelength range of the visible spectrum extends from violet (400 nm) to red (750 nm). Express these wavelengths in frequencies (Hz). ($1 \text{ nm} = 10^{-9} \text{ m}$)

Solution:

1. Identify Given Information:

• $\lambda_{\text{violet}} = 400 \text{ nm} = 400 \times 10^{-9} \text{ m}$
• $\lambda_{\text{red}} = 750 \text{ nm} = 750 \times 10^{-9} \text{ m}$
• $c = 3.00 \times 10^8 \text{ m s}^{-1}$

1. State the Formula: $$\nu = \frac{c}{\lambda}$$

2. Calculate Frequency for Red Light: $$\nu_{\text{red}} = \frac{3.00 \times 10^8 \text{ m s}^{-1}}{750 \times 10^{-9} \text{ m}}$$ $$\nu_{\text{red}} = \frac{3.00}{750} \times 10^{17} \text{ s}^{-1}$$ $$\nu_{\text{red}} = 0.004 \times 10^{17} \text{ Hz} = 4.00 \times 10^{14} \text{ Hz}$$

3. Calculate Frequency for Violet Light: $$\nu_{\text{violet}} = \frac{3.00 \times 10^8 \text{ m s}^{-1}}{400 \times 10^{-9} \text{ m}}$$ $$\nu_{\text{violet}} = \frac{3.00}{400} \times 10^{17} \text{ s}^{-1}$$ $$\nu_{\text{violet}} = 0.0075 \times 10^{17} \text{ Hz} = 7.50 \times 10^{14} \text{ Hz}$$

4. Final Answer: The range of the visible spectrum is from $4.0 \times 10^{14} \text{ Hz}$ to $7.5 \times 10^{14} \text{ Hz}$.

Takeaway: Notice that the longer wavelength (red) corresponds to the lower frequency, and the shorter wavelength (violet) corresponds to the higher frequency. This perfectly illustrates the inverse relationship between frequency and wavelength.

Example 3: Calculating Wavenumber and Frequency (Book Problem)

Calculate (a) wavenumber and (b) frequency of yellow radiation having wavelength 5800 $\text{\AA}$.

Solution:

1. Identify Given Information:

• Wavelength ($\lambda$) = $5800 \text{ \AA}$
• Conversion factor: $1 \text{ \AA} = 10^{-10} \text{ m} = 10^{-8} \text{ cm}$

1. Convert Wavelength to Standard Units:

• $\lambda = 5800 \times 10^{-8} \text{ cm}$
• $\lambda = 5800 \times 10^{-10} \text{ m}$

1. (a) Calculation of Wavenumber ($\bar{\nu}$):

• Formula: $\bar{\nu} = \frac{1}{\lambda}$
• Using wavelength in cm (as cm$^{-1}$ is the common unit for wavenumber): $$\bar{\nu} = \frac{1}{5800 \times 10^{-8} \text{ cm}}$$
• Alternatively, using meters first: $$\bar{\nu} = \frac{1}{5800 \times 10^{-10} \text{ m}} = 1.724 \times 10^6 \text{ m}^{-1}$$
• Converting to cm$^{-1}$: $$1.724 \times 10^6 \text{ m}^{-1} \times \frac{1 \text{ m}}{100 \text{ cm}} = 1.724 \times 10^4 \text{ cm}^{-1}$$

1. (b) Calculation of Frequency ($\nu$):

• Formula: $\nu = \frac{c}{\lambda}$
• Using wavelength in meters: $$\nu = \frac{3 \times 10^8 \text{ m s}^{-1}}{5800 \times 10^{-10} \text{ m}}$$ $$\nu = \frac{3}{5800} \times 10^{18} \text{ s}^{-1}$$ $$\nu = 5.172 \times 10^{14} \text{ s}^{-1} \text{ (or Hz)}$$

Takeaway: Wavenumber is most commonly expressed in $\text{cm}^{-1}$ in spectroscopy. Always be careful with Angstrom to cm and Angstrom to meter conversions.

Example 4: Energy of a Quantum of Red Light

Calculate the energy of one quantum of red light having a wavelength of 750 nm. (Given: $h = 6.626 \times 10^{-34} \text{ J s}$, $c = 3.0 \times 10^8 \text{ m s}^{-1}$)

Solution:

1. Identify Given Information:

• $\lambda = 750 \text{ nm} = 750 \times 10^{-9} \text{ m}$
• $h = 6.626 \times 10^{-34} \text{ J s}$
• $c = 3.0 \times 10^8 \text{ m s}^{-1}$

1. State the Formulas:

• Planck's quantum equation: $E = h\nu$
• Wave equation: $\nu = \frac{c}{\lambda}$
• Combining them: $E = \frac{hc}{\lambda}$

1. Perform the Calculation: $$E = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3.0 \times 10^8 \text{ m s}^{-1})}{750 \times 10^{-9} \text{ m}}$$

   First, multiply the constants in the numerator: $$hc = 19.878 \times 10^{-26} \text{ J m}$$

   Now, divide by the wavelength: $$E = \frac{19.878 \times 10^{-26}}{750 \times 10^{-9}}$$ $$E = 0.026504 \times 10^{-17} \text{ J}$$ $$E = 2.65 \times 10^{-19} \text{ J}$$

2. Verification: The energy of visible light photons is typically in the order of $10^{-19}$ Joules. The result matches this expected order of magnitude.

Takeaway: The combined formula $E = \frac{hc}{\lambda}$ is highly efficient for calculating energy directly from wavelength. Memorizing the approximate value of $hc \approx 19.88 \times 10^{-26} \text{ J m}$ can speed up calculations in competitive exams.

Example 5: Energy of a Quantum of Violet Light

Calculate the energy of one quantum of violet light having a frequency of $7.5 \times 10^{14} \text{ Hz}$. (Given: $h = 6.626 \times 10^{-34} \text{ J s}$)

Solution:

1. Identify Given Information:

• $\nu = 7.5 \times 10^{14} \text{ Hz} = 7.5 \times 10^{14} \text{ s}^{-1}$
• $h = 6.626 \times 10^{-34} \text{ J s}$

1. State the Formula: According to Planck's quantum theory, the energy of a quantum is directly proportional to its frequency: $$E = h\nu$$

2. Perform the Calculation: $$E = (6.626 \times 10^{-34} \text{ J s}) \times (7.5 \times 10^{14} \text{ s}^{-1})$$

   Multiply the numerical values: $$6.626 \times 7.5 = 49.695$$

   Multiply the powers of 10: $$10^{-34} \times 10^{14} = 10^{-20}$$

   Combine them: $$E = 49.695 \times 10^{-20} \text{ J}$$ $$E = 4.97 \times 10^{-19} \text{ J}$$

3. Comparison: In Example 4, the energy of red light (lower frequency) was $2.65 \times 10^{-19} \text{ J}$. Here, the energy of violet light (higher frequency) is $4.97 \times 10^{-19} \text{ J}$. This confirms that higher frequency radiation carries more energy per quantum.

Takeaway: Energy and frequency are directly proportional. Violet light is more energetic than red light.

Example 6: Calculating Wavelength from Energy

A particular electromagnetic radiation has a quantum energy of $3.313 \times 10^{-19} \text{ J}$. Calculate its wavelength in nanometers (nm). (Given: $h = 6.626 \times 10^{-34} \text{ J s}$, $c = 3.0 \times 10^8 \text{ m s}^{-1}$)

Solution:

1. Identify Given Information:

• $E = 3.313 \times 10^{-19} \text{ J}$
• $h = 6.626 \times 10^{-34} \text{ J s}$
• $c = 3.0 \times 10^8 \text{ m s}^{-1}$

1. State the Formula: $$E = \frac{hc}{\lambda}$$ Rearranging to solve for wavelength ($\lambda$): $$\lambda = \frac{hc}{E}$$

2. Perform the Calculation: $$\lambda = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3.0 \times 10^8 \text{ m s}^{-1})}{3.313 \times 10^{-19} \text{ J}}$$

   Notice that $6.626$ is exactly double $3.313$: $$\frac{6.626}{3.313} = 2$$

   Substitute this simplification back into the equation: $$\lambda = 2 \times 10^{-34} \times 3.0 \times 10^8 \times 10^{19} \text{ m}$$ $$\lambda = 6.0 \times 10^{-34 + 8 + 19} \text{ m}$$ $$\lambda = 6.0 \times 10^{-7} \text{ m}$$

3. Convert to Nanometers: $$1 \text{ nm} = 10^{-9} \text{ m}$$ $$\lambda = 6.0 \times 10^{-7} \text{ m} \times \frac{10^9 \text{ nm}}{1 \text{ m}}$$ $$\lambda = 6.0 \times 10^2 \text{ nm} = 600 \text{ nm}$$

Takeaway: Look for simple mathematical cancellations in competitive exams. The examiners often design problems where numbers like $6.626$ and $3.313$ cancel out neatly.

Example 7: Calculating Frequency from Wavenumber

The wavenumber of a specific infrared radiation is $1.0 \times 10^4 \text{ cm}^{-1}$. Calculate its frequency in Hz. (Given: $c = 3.0 \times 10^8 \text{ m s}^{-1}$)

Solution:

1. Identify Given Information:

• Wavenumber ($\bar{\nu}$) = $1.0 \times 10^4 \text{ cm}^{-1}$
• $c = 3.0 \times 10^8 \text{ m s}^{-1}$

1. Convert Wavenumber to SI Units (m$^{-1}$):

• $\bar{\nu} = 1.0 \times 10^4 \text{ cm}^{-1}$
• Since $1 \text{ cm} = 10^{-2} \text{ m}$, then $1 \text{ cm}^{-1} = (10^{-2} \text{ m})^{-1} = 10^2 \text{ m}^{-1}$
• $\bar{\nu} = 1.0 \times 10^4 \times 10^2 \text{ m}^{-1} = 1.0 \times 10^6 \text{ m}^{-1}$

1. State the Formulas:

• $\bar{\nu} = \frac{1}{\lambda} \implies \lambda = \frac{1}{\bar{\nu}}$
• $\nu = \frac{c}{\lambda}$
• Combining them: $\nu = c \times \bar{\nu}$

1. Perform the Calculation: $$\nu = (3.0 \times 10^8 \text{ m s}^{-1}) \times (1.0 \times 10^6 \text{ m}^{-1})$$ $$\nu = 3.0 \times 10^{14} \text{ s}^{-1} = 3.0 \times 10^{14} \text{ Hz}$$

Takeaway: The formula $\nu = c \bar{\nu}$ is a direct and fast way to convert wavenumber to frequency. However, you must ensure that $c$ and $\bar{\nu}$ use the same unit of length (meters) before multiplying.

Example 8: Comparing Energies of Different Wavelengths

Two electromagnetic radiations have wavelengths of $2000 \text{ \AA}$ and $4000 \text{ \AA}$ respectively. What is the ratio of their quantum energies?

Solution:

1. Identify Given Information:

• $\lambda_1 = 2000 \text{ \AA}$
• $\lambda_2 = 4000 \text{ \AA}$

1. State the Formula: The energy of a quantum is inversely proportional to its wavelength: $$E = \frac{hc}{\lambda}$$ Therefore, $E \propto \frac{1}{\lambda}$

2. Set Up the Ratio: $$\frac{E_1}{E_2} = \frac{\frac{hc}{\lambda_1}}{\frac{hc}{\lambda_2}} = \frac{\lambda_2}{\lambda_1}$$

3. Perform the Calculation: $$\frac{E_1}{E_2} = \frac{4000 \text{ \AA}}{2000 \text{ \AA}}$$ $$\frac{E_1}{E_2} = \frac{2}{1}$$

4. Conclusion: The ratio of their energies is 2:1. The radiation with half the wavelength has twice the energy.

Takeaway: For ratio problems, you do not need to plug in the values of $h$ and $c$, nor do you need to convert units to meters, as long as both wavelengths are in the same unit. This saves immense time in competitive exams.

Example 9: Wavenumber of a Microwave

A microwave used in radar has a frequency of $3.0 \times 10^{10} \text{ Hz}$. Calculate its wavenumber in $\text{m}^{-1}$. (Given: $c = 3.0 \times 10^8 \text{ m s}^{-1}$)

Solution:

1. Identify Given Information:

• $\nu = 3.0 \times 10^{10} \text{ Hz} = 3.0 \times 10^{10} \text{ s}^{-1}$
• $c = 3.0 \times 10^8 \text{ m s}^{-1}$

1. State the Formulas:

• $\lambda = \frac{c}{\nu}$
• $\bar{\nu} = \frac{1}{\lambda}$
• Combining them: $\bar{\nu} = \frac{\nu}{c}$

1. Perform the Calculation: $$\bar{\nu} = \frac{3.0 \times 10^{10} \text{ s}^{-1}}{3.0 \times 10^8 \text{ m s}^{-1}}$$ $$\bar{\nu} = 1.0 \times 10^2 \text{ m}^{-1}$$ $$\bar{\nu} = 100 \text{ m}^{-1}$$

2. Verification: Let's check the wavelength first. $\lambda = \frac{3.0 \times 10^8}{3.0 \times 10^{10}} = 10^{-2} \text{ m} = 1 \text{ cm}$. The wavenumber is $\frac{1}{\lambda} = \frac{1}{10^{-2} \text{ m}} = 100 \text{ m}^{-1}$. The calculation is correct.

Takeaway: The direct relation $\bar{\nu} = \frac{\nu}{c}$ is another handy derivation to keep in your memory capsule for quick problem-solving.

Example 10: Energy of an Ultraviolet Quantum

Calculate the energy of a quantum of ultraviolet radiation with a frequency of $1.0 \times 10^{16} \text{ Hz}$. (Given: $h = 6.626 \times 10^{-34} \text{ J s}$)

Solution:

1. Identify Given Information:

• $\nu = 1.0 \times 10^{16} \text{ Hz} = 1.0 \times 10^{16} \text{ s}^{-1}$
• $h = 6.626 \times 10^{-34} \text{ J s}$

1. State the Formula: $$E = h\nu$$

2. Perform the Calculation: $$E = (6.626 \times 10^{-34} \text{ J s}) \times (1.0 \times 10^{16} \text{ s}^{-1})$$ $$E = 6.626 \times 10^{-34 + 16} \text{ J}$$ $$E = 6.626 \times 10^{-18} \text{ J}$$

3. Comparison with Visible Light: Recall from previous examples that visible light energies are in the order of $10^{-19} \text{ J}$. This UV quantum has an energy of $\sim 6.6 \times 10^{-18} \text{ J}$, which is roughly 10 to 20 times more energetic than visible light. This aligns with the fact that UV radiation has a higher frequency than visible light.

Takeaway: Always do a quick mental check of the order of magnitude. UV radiation should have a higher energy exponent (less negative) than visible radiation.