Problem 58 - A Complete Resource Book in Chemistry for JEE Main

Question

The value of Planck constant is \(6.63 \times 10^{-34} \mathrm{Js}\). he velocity of light is \(3.0 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1} .\) which value is closest to the wavelength in nanometres of a quantum of light with frequency of \(8 \times 10^{15} \mathrm{~s}^{-1}\) ? (a) \(4 \times 10^{1}\) (b) \(3 \times 10^{7}\) (c) \(2 \times 10^{-25}\) (d) \(5 \times 10^{-18}\)

Step-by-Step Solution

Answer

The closest wavelength value is (a) \(4 \times 10^{1} = 40\) nm.

Step 1: Understand the Relationships

The relationship between the speed of light, frequency, and wavelength is given by the formula \( c = \lambda \times f \), where \( c \) is the speed of light, \( \lambda \) is the wavelength, and \( f \) is the frequency.

Step 2: Rearrange the Formula

To find the wavelength \( \lambda \), we rearrange the formula \( c = \lambda \times f \) to \( \lambda = \frac{c}{f} \).

Step 3: Plug in the Known Values

Substitute the given values into the formula: Speed of light \( c = 3.0 \times 10^8 \mathrm{~m/s} \) and frequency \( f = 8 \times 10^{15} \mathrm{~s^{-1}} \). So, \( \lambda = \frac{3.0 \times 10^8}{8 \times 10^{15}} \).

Step 4: Calculate Wavelength

Perform the calculation: \( \lambda = \frac{3.0 \times 10^8}{8 \times 10^{15}} = 3.75 \times 10^{-8} \mathrm{~m} \).

Step 5: Convert Meters to Nanometres

Since 1 meter = \(10^9\) nanometres, convert the result from meters to nanometres: \( 3.75 \times 10^{-8} \mathrm{~m} = 3.75 \times 10^{-8} \times 10^9 \mathrm{~nm} = 37.5 \mathrm{~nm} \).

Step 6: Find the Closest Answer

Compare the calculated wavelength (37.5 nm) with the provided options. The closest value is option (a) \(4 \times 10^{1} = 40 \mathrm{~nm}\).

Key Concepts

Planck constant

The Planck constant is a fundamental constant in physics. It symbolizes the link between the energy of a photon and the frequency of its electromagnetic wave. The value is approximately \(6.63 \times 10^{-34} \text{ Js}\). This tiny number indicates how small the energy of a single photon can be for typical frequencies.
The Planck constant plays a crucial role in quantum mechanics. It's central to the famous equation \(E = hf\), connecting energy (\(E\)) to frequency (\(f\)), with \(h\) being the Planck constant. This equation tells us that as the frequency of light increases, its energy does too.
Here are a few key points about the Planck constant:

• It's fundamental to understanding the nature of light and other electromagnetic waves.
• It emphasizes that energy comes in discrete units, known as quanta.
• It appears in various other equations, impacting numerous areas in physics.

Understanding this constant helps bridge the gap between classical and quantum physics, providing insights into how energy behaves at tiny scales.

speed of light

The speed of light in a vacuum is a huge number: \(3.0 \times 10^8 \text{ m/s}\). This is one of the most important constants in physics. Light is incredibly fast; it can circle the Earth more than seven times in just one second!
The speed of light is constant across the universe. This provides a backbone for many physics laws, like Einstein's theory of relativity, which relies heavily on this constant. The formula \(c = \lambda \times f\) ties together light's speed, its wavelength (\(\lambda\)), and its frequency (\(f\)).
Some key attributes about the speed of light include:

• It's the maximum speed at which information or energy can be transferred.
• It remains constant across different mediums, albeit a bit slower in materials like water or glass.
• This constancy is critical in calculating both wavelength and frequency for electromagnetic waves.

Having a grasp on light's speed aids in understanding numerous scientific phenomena, making it a cornerstone of modern physics.

frequency-wavelength relation

The frequency-wavelength relation is a simple yet crucial formula. It connects the speed of light \(c\), frequency \(f\), and wavelength \(\lambda\) through the equation \(c = \lambda \times f\). This means knowing any two of these values lets you find the third.
To find the wavelength of a light quantum, or photon, with a given frequency, you rearrange the relation to \(\lambda = \frac{c}{f}\). This calculation is essential in determining properties of light and other electromagnetic waves.
Here’s how it works:

• \(\lambda\) (wavelength) and \(f\) (frequency) are inversely related: larger wavelengths mean lower frequencies, and vice versa.
• This relation is fundamental in various fields like radio waves, microwaves, and even gamma rays.
• It shows how high-frequency waves, like X-rays, have much shorter wavelengths compared to visible light or radio waves.

Understanding this relationship helps explain how different types of electromagnetic waves behave and interact with matter. It plays a pivotal role in telecommunications, astronomy, and physics research.